MATH 132 - CALCULUS AND
ANALYTIC GEOMETRY II -
Integration by Parts
Question Bank - Set 1
Liberty University
Question 1
Question
Evaluate the integral ∫x·ln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫x·ln(x)dx using integration by parts, we’ll
choose u= ln(x)and dv =x dx. Then we’ll differentiate uto get du and
integrate dv to get v.
Step 1: Choose uand dv. Let u= ln(x)and dv =x dx.
Step 2: Calculate du and v. Differentiate uwith respect to xto get du:
du =1
xdx
Integrate dv with respect to xto get v:
v=1
2x2
Step 3: Apply integration by parts formula. The integration by parts
formula states: ∫u dv =uv −∫v du
Step 4: Substitute into the formula.
∫x·ln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
Step 5: Simplify the integral. Simplify the above expression to get:
1
2x2ln(x)−1
2∫x dx
Step 6: Evaluate the remaining integral. Now, integrate the remaining
term: ∫x dx =1
2x2
Step 7: Final answer. Substitute this back into the expression to get:
1
2x2ln(x)−1
4x2+C
where Cis the constant of integration.
Question 2
Question
Evaluate the integral ∫xcos(x)dx using integration by parts.
Solution
To evaluate the integral ∫xcos(x)dx using integration by parts, we will use the
formula ∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
Step 1: Let’s choose u=xand dv = cos(x)dx. Then, we have:
du =dx and v=∫cos(x)dx = sin(x).
Step 2: Now we can apply the integration by parts formula:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C,
where Cis the constant of integration.
Therefore, the integral ∫xcos(x)dx evaluates to xsin(x) + cos(x) + C.
2
Question 3
Question
Calculate the following integral using integration by parts:
∫x2ln(x)dx
Solution
To evaluate the integral ∫x2ln(x)dx, we’ll use integration by parts. Recall the
formula for integration by parts:
∫u dv =uv −∫v du
where uand vare differentiable functions of x. Let’s choose u= ln(x)and
dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Step 1: Calculate du and v:
du =1
xdx and v=1
3x3
Step 2: Apply the formula for integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 3: Evaluate the last integral:
∫x2dx =1
3x3+C
Step 4: Substitute the result back into the formula:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
Therefore, the integral ∫x2ln(x)dx simplifies to 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
Question 4
Question
Evaluate the integral ∫xexdx using integration by parts.
3
Solution
To evaluate the integral ∫xexdx using integration by parts, we will apply the
formula ∫u dv =uv −∫v du with u=xand dv =exdx.
Step 1: Choose uand dv. Let u=xand dv =exdx.
Step 2: Compute du and v. Compute du by taking the derivative of uwith
respect to x:du =dx
Integrate dv to find v:v=∫exdx =ex
Step 3: Apply the integration by parts formula. We have: ∫xexdx =
uv −∫v du
=x·ex−∫exdx
Step 4: Evaluate the integral. Continuing from where we left off: =x·ex−
∫exdx
=x·ex−ex+C
Therefore, ∫xexdx =x·ex−ex+C, where Cis the constant of integration.
Question 5
Question
Evaluate the integral ∫xln(x)dx by using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let u= ln(x)and dv =x dx. Then, du =1
xdx and v=1
2x2.
Step 2: Now, we can apply the integration by parts formula:
∫xln(x)dx =∫u dv
=uv −∫v du
= ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
4
Question 6
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To solve this integral, we will use integration by parts, which states that ∫u dv =
uv −∫v du. In this case, we will let u= ln(x)and dv =x2dx.
Step 1: Find du and v.
•u= ln(x)
•du =1
xdx
•v=∫x2dx =x3
3
Step 2: Apply integration by parts formula.
∫x2ln(x)dx =uv −∫v du
= ln(x)·x3
3−∫x3
3·1
xdx
=x3ln(x)
3−1
3∫x2dx
Step 3: Integrate the remaining integral.
∫x2ln(x)dx =x3ln(x)
3−1
3∫x2dx
=x3ln(x)
3−1
3·x3
3+C
=x3ln(x)
3−x3
9+C
Therefore, the solution to the integral ∫x2ln(x)dx is x3ln(x)
3−x3
9+C, where
Cis the constant of integration.
Question 7
Question
Evaluate the integral ∫xexdx using integration by parts.
5
Solution
To evaluate the integral ∫xexdx using integration by parts, we will apply the
formula ∫u dv =uv −∫v du.
Step 1: We will choose uand dv. Let u=xand dv =exdx.
Step 2: Calculate du and v.
•du =dx
• To find v, we integrate dv:
∫exdx =ex+C
So, v=ex
Step 3: Apply the integration by parts formula.
∫xexdx =uv −∫v du
=x·ex−∫exdx
Step 4: Simplify the integral.
=x·ex−ex+C= (x−1)ex+C
Thus, ∫xexdx = (x−1)ex+C.
Question 8
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫x2ln(x)dx using integration by parts, we will
choose u= ln(x)and dv =x2dx.
Step 1: Calculate du and v.
•u= ln(x)
•du =1
xdx
•v=∫x2dx =1
3x3
6
Step 2: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C, where Cis the constant of
integration.
Question 9
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate ∫xln(x)dx, we will use integration by parts. Recall the formula
for integration by parts: ∫u dv =uv −∫v du.
Step 1: Let us choose: u= ln(x)
dv =x dx
Differentiating uand integrating dv:du =1
xdx
v=1
2x2
Step 2: Write out the formula ∫u dv =uv −∫v du:∫xln(x)dx =
ln(x)·1
2x2−∫1
2x2·1
xdx
Simplify: =1
2x2ln(x)−1
2∫x dx
Step 3: Now, we integrate ∫x dx:=1
2x2ln(x)−1
2·1
2x2+C
Finally, simplifying the expression gives: =1
2x2ln(x)−1
4x2+C, where C
is the constant of integration.
7
Question 10
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate ∫x2exdx using integration by parts, we will use the formula
∫u dv =uv −∫v du where uand dv are functions of x.
Step 1: Let u=x2and dv =exdx. Then, we have:
du = 2x dx and v=∫exdx =ex
Step 2: Applying the integration by parts formula, we have:
∫x2exdx =x2ex−∫2xexdx
=x2ex−2∫xexdx
Step 3: Let’s now evaluate the remaining integral ∫xexdx. We again use
integration by parts with u=xand dv =exdx:
du =dx and v=∫exdx =ex
Step 4: Applying the integration by parts formula to the integral ∫xexdx,
we get:
∫xexdx =xex−∫exdx
=xex−ex
Step 5: Substituting this back into our original integral, we have:
∫x2exdx =x2ex−2∫xexdx
=x2ex−2(xex−ex)
=x2ex−2xex+ 2ex+C
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C, where Cis the constant of
integration.
8
Question 11
Question
Evaluate the integral ∫xcos(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xcos(x)dx using integration by parts, we will
apply the formula
∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
Step 1: Let’s choose u=xand dv = cos(x)dx. Then, we have:
du =dx
v=∫cos(x)dx = sin(x)
Step 2: Use the integration by parts formula:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C,
where Cis the constant of integration.
Hence, the result of the given integral is ∫xcos(x)dx =xsin(x)+cos(x)+C.
Question 12
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate the given integral ∫x2exdx using integration by parts, we will let
u=x2and dv =exdx. Then, we will find du and vas follows:
•du =d
dx (x2)dx = 2x dx
•v=∫exdx =ex
9
Now, we can apply the integration by parts formula:
∫u dv =uv −∫v du
Step 1: Apply the integration by parts formula.
∫x2exdx =x2ex−∫ex·2x dx
=x2ex−2∫xexdx
Now, we need to evaluate the remaining integral ∫xexdx using integration
by parts again.
Let u=xand dv =exdx. Then, find du and v:
•du =d
dx (x)dx =dx
•v=∫exdx =ex
Step 2: Apply integration by parts to the integral ∫xexdx.
∫xexdx =xex−∫exdx
=xex−ex
Now, substitute this result back into our original integral to solve for the
final answer.
Step 3: Substitute the result back into the original integral.
∫x2exdx =x2ex−2 (xex−ex)
=x2ex−2xex+ 2ex+C
So, ∫x2exdx =x2ex−2xex+2ex+C, where Cis the constant of integration.
Question 13
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To find the integral of x2ex, we will use integration by parts. Let’s choose our
functions uand dv such that du is easy to compute and vis easy to differentiate.
Let u=x2and dv =exdx. Then, we have:
10
] Compute du and v.
du = 2x dx
To find v, integrate dv with respect to x.
v=∫exdx =ex] Apply integration by parts formula:
∫u dv =uv −∫v du
This gives us:
∫x2exdx =x2ex−∫2xexdx
] Integrate the remaining integral by parts: Let u=xand dv = 2exdx.
Compute du and vas follows:
du =dx
v=∫2exdx = 2ex
] Apply integration by parts formula again:
∫2xexdx = 2xex−∫2exdx
Substitute back into the previous equation:
∫x2exdx =x2ex−(2xex−2ex)
∫x2exdx =x2ex−2xex+ 2ex+C
where Cis the constant of integration.
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C.
Question 14
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will apply
the formula: ∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
11
Step 1: Let’s choose u= ln(x)and dv =x2dx. Then, du =1
xdx and
v=1
3x3.
Step 2: Now, let’s apply the integration by parts formula:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C,
where Cis the constant of integration.
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C.
Question 15
Question
Evaluate the definite integral ∫1
0xsin−1(x)dx using integration by parts.
Solution
To evaluate the given definite integral ∫1
0xsin−1(x)dx using integration by
parts, we will use the formula ∫u dv =uv −∫v du.
Step 1: Choose uand dv
Let u= sin−1(x)and dv =x dx. Then, du =1
√1−x2dx and v=1
2x2.
Step 2: Apply integration by parts
∫xsin−1(x)dx =uv −∫v du
= sin−1(x)(1
2x2)−∫(1
2x2·1
√1−x2)dx
=1
2x2sin−1(x)−1
2∫x2
√1−x2dx
Step 3: Evaluate the remaining integral
To evaluate ∫x2
√1−x2dx, we can use the substitution u= 1−x2,du =−2x dx:
∫x2
√1−x2dx =−1
2∫−2x
√1−x2·x dx
=−1
2∫−du
√u
=√u+C
=√1−x2+C
12
Step 4: Substitute back and apply limits
∫1
0
xsin−1(x)dx =[1
2x2sin−1(x)−1
2(√1−x2)]1
0
=(1
2sin−1(1) −1
2√1−1)−(1
2sin−1(0) −1
2√1−0)
=π
4
Therefore, ∫1
0xsin−1(x)dx =π
4.
Question 16
Question
Evaluate the integral: ∫x2exdx.
Solution
To evaluate the given integral, we will use integration by parts. Integration by
parts is given by the formula:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let u=x2and dv =exdx. Then, calculate du and v:
du = 2x dx and v=∫exdx =ex
Step 2: Apply the formula for integration by parts:
∫x2exdx =x2ex−∫2xexdx
Step 3: Now, we need to evaluate the remaining integral on the right-hand
side, ∫2xexdx, using integration by parts again.
Step 4: Let u= 2xand dv =exdx. Calculate du and v:
du = 2 dx and v=ex
Step 5: Apply the formula for integration by parts to evaluate ∫2xexdx:
∫2xexdx = 2xex−∫2exdx
13
Step 6: Simplify the integral:
∫2exdx = 2 ∫exdx = 2ex
Step 7: Substitute back into our original integral:
∫x2exdx =x2ex−2xex+C
Therefore, ∫x2exdx =x2ex−2xex+C, where Cis the constant of integra-
tion.
Question 17
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx, we can use the integration by parts for-
mula: ∫u dv =uv −∫v du.
Let u= ln(x)and dv =x dx. Then, we have:
du =1
xdx
v=1
2x2
14
Now, we will apply the integration by parts formula:
Step 1: Let u= ln(x), dv =x dx
Step 2: Calculate du and v:
du =1
xdx, v =1
2x2
Step 3: Apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)(1
2x2)−∫(1
2x2)(1
x)dx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 18
Question
Compute the integral ∫x2exdx using integration by parts.
Solution
To evaluate the given integral ∫x2exdx using integration by parts, we will apply
the formula ∫u dv =uv −∫v du with u=x2and dv =exdx.
Step 1: Let’s determine du and v. We have:
du =d
dx (x2)dx = 2x dx
v=∫exdx =ex
Step 2: Apply integration by parts formula.
∫x2exdx =x2ex−∫ex·2x dx
=x2ex−2∫xexdx
15
Now, we need to apply integration by parts again to evaluate the remaining
integral ∫xexdx.
Step 3: Let’s apply integration by parts to ∫xexdx. Choose u=xand
dv =exdx.
Now we determine du and v:
du =dx
v=∫exdx =ex
Step 4: Apply integration by parts formula to ∫xexdx.
∫xexdx =xex−∫exdx
=xex−ex
Step 5: Substitute back into our original integral.
∫x2exdx =x2ex−2∫xexdx
=x2ex−2(xex−ex)
=x2ex−2xex+ 2ex+C
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C, where Cis the constant of
integration.
Question 19
Question
Evaluate the following integral:
∫x2cos x dx
Solution
To evaluate the given integral ∫x2cos x dx, we will use integration by parts.
Step 1: Let’s choose our uand dv. Let u=x2and dv = cos x dx.
Step 2: Compute du and v.
•du =d
dx (x2)dx = 2x dx
•v=∫cos x dx = sin x
16
Step 3: Apply the formula for integration by parts.
∫u dv =uv −∫v du
Now, we substitute u,dv,du, and vinto the formula:
∫x2cos x dx =x2sin x−∫sin x·2x dx
Step 4: Simplify the resulting integral.
∫x2cos x dx =x2sin x−2∫xsin x dx
Step 5: Let’s use integration by parts again for the remaining integral. Let
u=xand dv = sin x dx.
Now, compute du and v:
•du =d
dx (x)dx =dx
•v=∫sin x dx =−cos x
Step 6: Apply the formula for integration by parts again.
∫x2cos x dx =x2sin x−2(x(−cos x)−∫(−cos x)dx)
Simplify further to get the final result:
∫x2cos x dx =x2sin x+ 2xcos x−2 sin x+C
Where Cis the constant of integration.
Question 20
Question
Evaluate the integral ∫x2sin−1x dx using integration by parts.
Solution
To evaluate the given integral ∫x2sin−1x dx using integration by parts, we need
to choose two functions to assign as uand dv in the formula:
∫u dv =uv −∫v du
Let’s choose u= sin−1xand dv =x2dx. Then, we can find du and v:
du =1
√1−x2dx
17
v=1
3x3
Now, we can apply the integration by parts formula:
∫x2sin−1x dx =1
3x3sin−1x−∫1
3x3·1
√1−x2dx
∫x2sin−1x dx =1
3x3sin−1x−1
3∫x3
√1−x2dx
Integrating the remaining integral ∫x3
√1−x2dx would require further manip-
ulation using substitution or other techniques.
Question 21
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x2ln(x)dx, we will use integration by parts, which
states ∫u dv =uv −∫v du.
Let u= ln(x)and dv =x2dx. Then, we have
du =1
xdx and v=1
3x3.
Step 1: Apply integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx.
Step 2: Integrate the remaining integral:
∫x2dx =1
3x3+C,
where Cis the constant of integration.
Step 3: Substitute back to find the final answer:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
=1
3x3(ln(x)−1
3) + C.
Therefore, ∫x2ln(x)dx =1
3x3(ln(x)−1
3) + C.
18
Question 22
Question
Evaluate the integral
∫xe2xdx
Solution
To evaluate the given integral, we will use the technique of integration by parts,
which states ∫u dv =uv −∫v du
Let u=xand dv =e2xdx. Then, we have du =dx and v=1
2e2x.
Step 1: Apply integration by parts:
∫xe2xdx =x(1
2e2x)−∫(1
2e2x)dx
=1
2xe2x−1
4e2x+C
Step 2: Simplify the result:
∫xe2xdx =1
2xe2x−1
4e2x+C
Therefore, the integral ∫xe2xdx evaluates to 1
2xe2x−1
4e2x+C.
Question 23
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will have to
choose which part of the integrand to differentiate and which part to integrate.
Let’s follow the formula for integration by parts:
∫u dv =uv −∫v du
Step 1: Choose u= ln(x)and dv =x dx.
Then, differentiate uto get:
du =1
xdx
19
And integrate dv to get:
v=x2
2
Step 2: Apply the integration by parts formula:
∫xln(x)dx = ln(x)·x2
2−∫x2
2·1
xdx
Simplify the expression:
∫xln(x)dx =x2
2ln(x)−1
2∫x dx
Step 3: Integrate the remaining integral:
∫x dx =x2
2
Step 4: Substitute back into the equation:
∫xln(x)dx =x2
2ln(x)−1
2·x2
2+C
∫xln(x)dx =x2
2ln(x)−x2
4+C
Therefore, the result of the given integral is x2
2ln(x)−x2
4+C, where Cis
the constant of integration.
Question 24
Question
Evaluate the integral ∫excos(x)dx using integration by parts.
Solution
To evaluate the given integral ∫excos(x)dx using integration by parts, we will
apply the formula:
∫u dv =uv −∫v du
Step 1: Let’s choose uand dv:
u=exand dv = cos(x)dx
Step 2: Compute du and v:
du =exdx
20
v=∫cos(x)dx = sin(x)
Step 3: Apply the integration by parts formula:
∫excos(x)dx =exsin(x)−∫sin(x)·exdx
Step 4: Now we have a new integral to evaluate. Let’s again apply integra-
tion by parts with u= sin(x)and dv =exdx:
u= sin(x)and dv =exdx
du = cos(x)dx
v=∫exdx =ex
Step 5: Apply the integration by parts formula:
∫sin(x)·exdx =exsin(x)−∫excos(x)dx
Step 6: Substitute this back into the first equation:
∫excos(x)dx =exsin(x)−(exsin(x)−∫excos(x)dx)
Step 7: Simplify and solve for the original integral:
2∫excos(x)dx =exsin(x)
∫excos(x)dx =exsin(x)
2+C
Therefore, ∫excos(x)dx =exsin(x)
2+C, where Cis the constant of integra-
tion.
Question 25
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
21
Solution
To integrate ∫xln(x)dx, we will use integration by parts where we will set
u= ln(x)and dv =x dx.
Step 1: Compute du and v.
•u= ln(x)
•du =1
xdx
•dv =x dx =⇒v=1
2x2
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C(where Cis the constant of integration)
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C.
22
Step 5: Simplify the integral. Simplify the above expression to get:
1
2x2ln(x)−1
2∫x dx
Step 6: Evaluate the remaining integral. Now, integrate the remaining
term: ∫x dx =1
2x2
Step 7: Final answer. Substitute this back into the expression to get:
1
2x2ln(x)−1
4x2+C
where Cis the constant of integration.
Question 2
Question
Evaluate the integral ∫xcos(x)dx using integration by parts.
Solution
To evaluate the integral ∫xcos(x)dx using integration by parts, we will use the
formula ∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
Step 1: Let’s choose u=xand dv = cos(x)dx. Then, we have:
du =dx and v=∫cos(x)dx = sin(x).
Step 2: Now we can apply the integration by parts formula:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C,
where Cis the constant of integration.
Therefore, the integral ∫xcos(x)dx evaluates to xsin(x) + cos(x) + C.
2
Question 3
Question
Calculate the following integral using integration by parts:
∫x2ln(x)dx
Solution
To evaluate the integral ∫x2ln(x)dx, we’ll use integration by parts. Recall the
formula for integration by parts:
∫u dv =uv −∫v du
where uand vare differentiable functions of x. Let’s choose u= ln(x)and
dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Step 1: Calculate du and v:
du =1
xdx and v=1
3x3
Step 2: Apply the formula for integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 3: Evaluate the last integral:
∫x2dx =1
3x3+C
Step 4: Substitute the result back into the formula:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
Therefore, the integral ∫x2ln(x)dx simplifies to 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
Question 4
Question
Evaluate the integral ∫xexdx using integration by parts.
3
Solution
To evaluate the integral ∫xexdx using integration by parts, we will apply the
formula ∫u dv =uv −∫v du with u=xand dv =exdx.
Step 1: Choose uand dv. Let u=xand dv =exdx.
Step 2: Compute du and v. Compute du by taking the derivative of uwith
respect to x:du =dx
Integrate dv to find v:v=∫exdx =ex
Step 3: Apply the integration by parts formula. We have: ∫xexdx =
uv −∫v du
=x·ex−∫exdx
Step 4: Evaluate the integral. Continuing from where we left off: =x·ex−
∫exdx
=x·ex−ex+C
Therefore, ∫xexdx =x·ex−ex+C, where Cis the constant of integration.
Question 5
Question
Evaluate the integral ∫xln(x)dx by using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let u= ln(x)and dv =x dx. Then, du =1
xdx and v=1
2x2.
Step 2: Now, we can apply the integration by parts formula:
∫xln(x)dx =∫u dv
=uv −∫v du
= ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
4
Question 6
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To solve this integral, we will use integration by parts, which states that ∫u dv =
uv −∫v du. In this case, we will let u= ln(x)and dv =x2dx.
Step 1: Find du and v.
•u= ln(x)
•du =1
xdx
•v=∫x2dx =x3
3
Step 2: Apply integration by parts formula.
∫x2ln(x)dx =uv −∫v du
= ln(x)·x3
3−∫x3
3·1
xdx
=x3ln(x)
3−1
3∫x2dx
Step 3: Integrate the remaining integral.
∫x2ln(x)dx =x3ln(x)
3−1
3∫x2dx
=x3ln(x)
3−1
3·x3
3+C
=x3ln(x)
3−x3
9+C
Therefore, the solution to the integral ∫x2ln(x)dx is x3ln(x)
3−x3
9+C, where
Cis the constant of integration.
Question 7
Question
Evaluate the integral ∫xexdx using integration by parts.
5
Solution
To evaluate the integral ∫xexdx using integration by parts, we will apply the
formula ∫u dv =uv −∫v du.
Step 1: We will choose uand dv. Let u=xand dv =exdx.
Step 2: Calculate du and v.
•du =dx
• To find v, we integrate dv:
∫exdx =ex+C
So, v=ex
Step 3: Apply the integration by parts formula.
∫xexdx =uv −∫v du
=x·ex−∫exdx
Step 4: Simplify the integral.
=x·ex−ex+C= (x−1)ex+C
Thus, ∫xexdx = (x−1)ex+C.
Question 8
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫x2ln(x)dx using integration by parts, we will
choose u= ln(x)and dv =x2dx.
Step 1: Calculate du and v.
•u= ln(x)
•du =1
xdx
•v=∫x2dx =1
3x3
6
Step 2: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C, where Cis the constant of
integration.
Question 9
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate ∫xln(x)dx, we will use integration by parts. Recall the formula
for integration by parts: ∫u dv =uv −∫v du.
Step 1: Let us choose: u= ln(x)
dv =x dx
Differentiating uand integrating dv:du =1
xdx
v=1
2x2
Step 2: Write out the formula ∫u dv =uv −∫v du:∫xln(x)dx =
ln(x)·1
2x2−∫1
2x2·1
xdx
Simplify: =1
2x2ln(x)−1
2∫x dx
Step 3: Now, we integrate ∫x dx:=1
2x2ln(x)−1
2·1
2x2+C
Finally, simplifying the expression gives: =1
2x2ln(x)−1
4x2+C, where C
is the constant of integration.
7
Question 10
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate ∫x2exdx using integration by parts, we will use the formula
∫u dv =uv −∫v du where uand dv are functions of x.
Step 1: Let u=x2and dv =exdx. Then, we have:
du = 2x dx and v=∫exdx =ex
Step 2: Applying the integration by parts formula, we have:
∫x2exdx =x2ex−∫2xexdx
=x2ex−2∫xexdx
Step 3: Let’s now evaluate the remaining integral ∫xexdx. We again use
integration by parts with u=xand dv =exdx:
du =dx and v=∫exdx =ex
Step 4: Applying the integration by parts formula to the integral ∫xexdx,
we get:
∫xexdx =xex−∫exdx
=xex−ex
Step 5: Substituting this back into our original integral, we have:
∫x2exdx =x2ex−2∫xexdx
=x2ex−2(xex−ex)
=x2ex−2xex+ 2ex+C
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C, where Cis the constant of
integration.
8
Question 11
Question
Evaluate the integral ∫xcos(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xcos(x)dx using integration by parts, we will
apply the formula
∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
Step 1: Let’s choose u=xand dv = cos(x)dx. Then, we have:
du =dx
v=∫cos(x)dx = sin(x)
Step 2: Use the integration by parts formula:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C,
where Cis the constant of integration.
Hence, the result of the given integral is ∫xcos(x)dx =xsin(x)+cos(x)+C.
Question 12
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate the given integral ∫x2exdx using integration by parts, we will let
u=x2and dv =exdx. Then, we will find du and vas follows:
•du =d
dx (x2)dx = 2x dx
•v=∫exdx =ex
9
Now, we can apply the integration by parts formula:
∫u dv =uv −∫v du
Step 1: Apply the integration by parts formula.
∫x2exdx =x2ex−∫ex·2x dx
=x2ex−2∫xexdx
Now, we need to evaluate the remaining integral ∫xexdx using integration
by parts again.
Let u=xand dv =exdx. Then, find du and v:
•du =d
dx (x)dx =dx
•v=∫exdx =ex
Step 2: Apply integration by parts to the integral ∫xexdx.
∫xexdx =xex−∫exdx
=xex−ex
Now, substitute this result back into our original integral to solve for the
final answer.
Step 3: Substitute the result back into the original integral.
∫x2exdx =x2ex−2 (xex−ex)
=x2ex−2xex+ 2ex+C
So, ∫x2exdx =x2ex−2xex+2ex+C, where Cis the constant of integration.
Question 13
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To find the integral of x2ex, we will use integration by parts. Let’s choose our
functions uand dv such that du is easy to compute and vis easy to differentiate.
Let u=x2and dv =exdx. Then, we have:
10
] Compute du and v.
du = 2x dx
To find v, integrate dv with respect to x.
v=∫exdx =ex] Apply integration by parts formula:
∫u dv =uv −∫v du
This gives us:
∫x2exdx =x2ex−∫2xexdx
] Integrate the remaining integral by parts: Let u=xand dv = 2exdx.
Compute du and vas follows:
du =dx
v=∫2exdx = 2ex
] Apply integration by parts formula again:
∫2xexdx = 2xex−∫2exdx
Substitute back into the previous equation:
∫x2exdx =x2ex−(2xex−2ex)
∫x2exdx =x2ex−2xex+ 2ex+C
where Cis the constant of integration.
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C.
Question 14
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will apply
the formula: ∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
11
Step 1: Let’s choose u= ln(x)and dv =x2dx. Then, du =1
xdx and
v=1
3x3.
Step 2: Now, let’s apply the integration by parts formula:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C,
where Cis the constant of integration.
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C.
Question 15
Question
Evaluate the definite integral ∫1
0xsin−1(x)dx using integration by parts.
Solution
To evaluate the given definite integral ∫1
0xsin−1(x)dx using integration by
parts, we will use the formula ∫u dv =uv −∫v du.
Step 1: Choose uand dv
Let u= sin−1(x)and dv =x dx. Then, du =1
√1−x2dx and v=1
2x2.
Step 2: Apply integration by parts
∫xsin−1(x)dx =uv −∫v du
= sin−1(x)(1
2x2)−∫(1
2x2·1
√1−x2)dx
=1
2x2sin−1(x)−1
2∫x2
√1−x2dx
Step 3: Evaluate the remaining integral
To evaluate ∫x2
√1−x2dx, we can use the substitution u= 1−x2,du =−2x dx:
∫x2
√1−x2dx =−1
2∫−2x
√1−x2·x dx
=−1
2∫−du
√u
=√u+C
=√1−x2+C
12
Step 4: Substitute back and apply limits
∫1
0
xsin−1(x)dx =[1
2x2sin−1(x)−1
2(√1−x2)]1
0
=(1
2sin−1(1) −1
2√1−1)−(1
2sin−1(0) −1
2√1−0)
=π
4
Therefore, ∫1
0xsin−1(x)dx =π
4.
Question 16
Question
Evaluate the integral: ∫x2exdx.
Solution
To evaluate the given integral, we will use integration by parts. Integration by
parts is given by the formula:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let u=x2and dv =exdx. Then, calculate du and v:
du = 2x dx and v=∫exdx =ex
Step 2: Apply the formula for integration by parts:
∫x2exdx =x2ex−∫2xexdx
Step 3: Now, we need to evaluate the remaining integral on the right-hand
side, ∫2xexdx, using integration by parts again.
Step 4: Let u= 2xand dv =exdx. Calculate du and v:
du = 2 dx and v=ex
Step 5: Apply the formula for integration by parts to evaluate ∫2xexdx:
∫2xexdx = 2xex−∫2exdx
13
Step 6: Simplify the integral:
∫2exdx = 2 ∫exdx = 2ex
Step 7: Substitute back into our original integral:
∫x2exdx =x2ex−2xex+C
Therefore, ∫x2exdx =x2ex−2xex+C, where Cis the constant of integra-
tion.
Question 17
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx, we can use the integration by parts for-
mula: ∫u dv =uv −∫v du.
Let u= ln(x)and dv =x dx. Then, we have:
du =1
xdx
v=1
2x2
14
Now, we will apply the integration by parts formula:
Step 1: Let u= ln(x), dv =x dx
Step 2: Calculate du and v:
du =1
xdx, v =1
2x2
Step 3: Apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)(1
2x2)−∫(1
2x2)(1
x)dx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 18
Question
Compute the integral ∫x2exdx using integration by parts.
Solution
To evaluate the given integral ∫x2exdx using integration by parts, we will apply
the formula ∫u dv =uv −∫v du with u=x2and dv =exdx.
Step 1: Let’s determine du and v. We have:
du =d
dx (x2)dx = 2x dx
v=∫exdx =ex
Step 2: Apply integration by parts formula.
∫x2exdx =x2ex−∫ex·2x dx
=x2ex−2∫xexdx
15
Now, we need to apply integration by parts again to evaluate the remaining
integral ∫xexdx.
Step 3: Let’s apply integration by parts to ∫xexdx. Choose u=xand
dv =exdx.
Now we determine du and v:
du =dx
v=∫exdx =ex
Step 4: Apply integration by parts formula to ∫xexdx.
∫xexdx =xex−∫exdx
=xex−ex
Step 5: Substitute back into our original integral.
∫x2exdx =x2ex−2∫xexdx
=x2ex−2(xex−ex)
=x2ex−2xex+ 2ex+C
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C, where Cis the constant of
integration.
Question 19
Question
Evaluate the following integral:
∫x2cos x dx
Solution
To evaluate the given integral ∫x2cos x dx, we will use integration by parts.
Step 1: Let’s choose our uand dv. Let u=x2and dv = cos x dx.
Step 2: Compute du and v.
•du =d
dx (x2)dx = 2x dx
•v=∫cos x dx = sin x
16
Step 3: Apply the formula for integration by parts.
∫u dv =uv −∫v du
Now, we substitute u,dv,du, and vinto the formula:
∫x2cos x dx =x2sin x−∫sin x·2x dx
Step 4: Simplify the resulting integral.
∫x2cos x dx =x2sin x−2∫xsin x dx
Step 5: Let’s use integration by parts again for the remaining integral. Let
u=xand dv = sin x dx.
Now, compute du and v:
•du =d
dx (x)dx =dx
•v=∫sin x dx =−cos x
Step 6: Apply the formula for integration by parts again.
∫x2cos x dx =x2sin x−2(x(−cos x)−∫(−cos x)dx)
Simplify further to get the final result:
∫x2cos x dx =x2sin x+ 2xcos x−2 sin x+C
Where Cis the constant of integration.
Question 20
Question
Evaluate the integral ∫x2sin−1x dx using integration by parts.
Solution
To evaluate the given integral ∫x2sin−1x dx using integration by parts, we need
to choose two functions to assign as uand dv in the formula:
∫u dv =uv −∫v du
Let’s choose u= sin−1xand dv =x2dx. Then, we can find du and v:
du =1
√1−x2dx
17
v=1
3x3
Now, we can apply the integration by parts formula:
∫x2sin−1x dx =1
3x3sin−1x−∫1
3x3·1
√1−x2dx
∫x2sin−1x dx =1
3x3sin−1x−1
3∫x3
√1−x2dx
Integrating the remaining integral ∫x3
√1−x2dx would require further manip-
ulation using substitution or other techniques.
Question 21
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x2ln(x)dx, we will use integration by parts, which
states ∫u dv =uv −∫v du.
Let u= ln(x)and dv =x2dx. Then, we have
du =1
xdx and v=1
3x3.
Step 1: Apply integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx.
Step 2: Integrate the remaining integral:
∫x2dx =1
3x3+C,
where Cis the constant of integration.
Step 3: Substitute back to find the final answer:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
=1
3x3(ln(x)−1
3) + C.
Therefore, ∫x2ln(x)dx =1
3x3(ln(x)−1
3) + C.
18
Question 22
Question
Evaluate the integral
∫xe2xdx
Solution
To evaluate the given integral, we will use the technique of integration by parts,
which states ∫u dv =uv −∫v du
Let u=xand dv =e2xdx. Then, we have du =dx and v=1
2e2x.
Step 1: Apply integration by parts:
∫xe2xdx =x(1
2e2x)−∫(1
2e2x)dx
=1
2xe2x−1
4e2x+C
Step 2: Simplify the result:
∫xe2xdx =1
2xe2x−1
4e2x+C
Therefore, the integral ∫xe2xdx evaluates to 1
2xe2x−1
4e2x+C.
Question 23
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will have to
choose which part of the integrand to differentiate and which part to integrate.
Let’s follow the formula for integration by parts:
∫u dv =uv −∫v du
Step 1: Choose u= ln(x)and dv =x dx.
Then, differentiate uto get:
du =1
xdx
19
And integrate dv to get:
v=x2
2
Step 2: Apply the integration by parts formula:
∫xln(x)dx = ln(x)·x2
2−∫x2
2·1
xdx
Simplify the expression:
∫xln(x)dx =x2
2ln(x)−1
2∫x dx
Step 3: Integrate the remaining integral:
∫x dx =x2
2
Step 4: Substitute back into the equation:
∫xln(x)dx =x2
2ln(x)−1
2·x2
2+C
∫xln(x)dx =x2
2ln(x)−x2
4+C
Therefore, the result of the given integral is x2
2ln(x)−x2
4+C, where Cis
the constant of integration.
Question 24
Question
Evaluate the integral ∫excos(x)dx using integration by parts.
Solution
To evaluate the given integral ∫excos(x)dx using integration by parts, we will
apply the formula:
∫u dv =uv −∫v du
Step 1: Let’s choose uand dv:
u=exand dv = cos(x)dx
Step 2: Compute du and v:
du =exdx
20
v=∫cos(x)dx = sin(x)
Step 3: Apply the integration by parts formula:
∫excos(x)dx =exsin(x)−∫sin(x)·exdx
Step 4: Now we have a new integral to evaluate. Let’s again apply integra-
tion by parts with u= sin(x)and dv =exdx:
u= sin(x)and dv =exdx
du = cos(x)dx
v=∫exdx =ex
Step 5: Apply the integration by parts formula:
∫sin(x)·exdx =exsin(x)−∫excos(x)dx
Step 6: Substitute this back into the first equation:
∫excos(x)dx =exsin(x)−(exsin(x)−∫excos(x)dx)
Step 7: Simplify and solve for the original integral:
2∫excos(x)dx =exsin(x)
∫excos(x)dx =exsin(x)
2+C
Therefore, ∫excos(x)dx =exsin(x)
2+C, where Cis the constant of integra-
tion.
Question 25
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
21
Solution
To integrate ∫xln(x)dx, we will use integration by parts where we will set
u= ln(x)and dv =x dx.
Step 1: Compute du and v.
•u= ln(x)
•du =1
xdx
•dv =x dx =⇒v=1
2x2
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C(where Cis the constant of integration)
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C.
22
Step 5: Simplify the integral. Simplify the above expression to get:
1
2x2ln(x)−1
2∫x dx
Step 6: Evaluate the remaining integral. Now, integrate the remaining
term: ∫x dx =1
2x2
Step 7: Final answer. Substitute this back into the expression to get:
1
2x2ln(x)−1
4x2+C
where Cis the constant of integration.
Question 2
Question
Evaluate the integral ∫xcos(x)dx using integration by parts.
Solution
To evaluate the integral ∫xcos(x)dx using integration by parts, we will use the
formula ∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
Step 1: Let’s choose u=xand dv = cos(x)dx. Then, we have:
du =dx and v=∫cos(x)dx = sin(x).
Step 2: Now we can apply the integration by parts formula:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C,
where Cis the constant of integration.
Therefore, the integral ∫xcos(x)dx evaluates to xsin(x) + cos(x) + C.
2
Question 3
Question
Calculate the following integral using integration by parts:
∫x2ln(x)dx
Solution
To evaluate the integral ∫x2ln(x)dx, we’ll use integration by parts. Recall the
formula for integration by parts:
∫u dv =uv −∫v du
where uand vare differentiable functions of x. Let’s choose u= ln(x)and
dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Step 1: Calculate du and v:
du =1
xdx and v=1
3x3
Step 2: Apply the formula for integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 3: Evaluate the last integral:
∫x2dx =1
3x3+C
Step 4: Substitute the result back into the formula:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
Therefore, the integral ∫x2ln(x)dx simplifies to 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
Question 4
Question
Evaluate the integral ∫xexdx using integration by parts.
3
Solution
To evaluate the integral ∫xexdx using integration by parts, we will apply the
formula ∫u dv =uv −∫v du with u=xand dv =exdx.
Step 1: Choose uand dv. Let u=xand dv =exdx.
Step 2: Compute du and v. Compute du by taking the derivative of uwith
respect to x:du =dx
Integrate dv to find v:v=∫exdx =ex
Step 3: Apply the integration by parts formula. We have: ∫xexdx =
uv −∫v du
=x·ex−∫exdx
Step 4: Evaluate the integral. Continuing from where we left off: =x·ex−
∫exdx
=x·ex−ex+C
Therefore, ∫xexdx =x·ex−ex+C, where Cis the constant of integration.
Question 5
Question
Evaluate the integral ∫xln(x)dx by using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let u= ln(x)and dv =x dx. Then, du =1
xdx and v=1
2x2.
Step 2: Now, we can apply the integration by parts formula:
∫xln(x)dx =∫u dv
=uv −∫v du
= ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
4
Question 6
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To solve this integral, we will use integration by parts, which states that ∫u dv =
uv −∫v du. In this case, we will let u= ln(x)and dv =x2dx.
Step 1: Find du and v.
•u= ln(x)
•du =1
xdx
•v=∫x2dx =x3
3
Step 2: Apply integration by parts formula.
∫x2ln(x)dx =uv −∫v du
= ln(x)·x3
3−∫x3
3·1
xdx
=x3ln(x)
3−1
3∫x2dx
Step 3: Integrate the remaining integral.
∫x2ln(x)dx =x3ln(x)
3−1
3∫x2dx
=x3ln(x)
3−1
3·x3
3+C
=x3ln(x)
3−x3
9+C
Therefore, the solution to the integral ∫x2ln(x)dx is x3ln(x)
3−x3
9+C, where
Cis the constant of integration.
Question 7
Question
Evaluate the integral ∫xexdx using integration by parts.
5
Solution
To evaluate the integral ∫xexdx using integration by parts, we will apply the
formula ∫u dv =uv −∫v du.
Step 1: We will choose uand dv. Let u=xand dv =exdx.
Step 2: Calculate du and v.
•du =dx
• To find v, we integrate dv:
∫exdx =ex+C
So, v=ex
Step 3: Apply the integration by parts formula.
∫xexdx =uv −∫v du
=x·ex−∫exdx
Step 4: Simplify the integral.
=x·ex−ex+C= (x−1)ex+C
Thus, ∫xexdx = (x−1)ex+C.
Question 8
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫x2ln(x)dx using integration by parts, we will
choose u= ln(x)and dv =x2dx.
Step 1: Calculate du and v.
•u= ln(x)
•du =1
xdx
•v=∫x2dx =1
3x3
6
Step 2: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C, where Cis the constant of
integration.
Question 9
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate ∫xln(x)dx, we will use integration by parts. Recall the formula
for integration by parts: ∫u dv =uv −∫v du.
Step 1: Let us choose: u= ln(x)
dv =x dx
Differentiating uand integrating dv:du =1
xdx
v=1
2x2
Step 2: Write out the formula ∫u dv =uv −∫v du:∫xln(x)dx =
ln(x)·1
2x2−∫1
2x2·1
xdx
Simplify: =1
2x2ln(x)−1
2∫x dx
Step 3: Now, we integrate ∫x dx:=1
2x2ln(x)−1
2·1
2x2+C
Finally, simplifying the expression gives: =1
2x2ln(x)−1
4x2+C, where C
is the constant of integration.
7
Question 10
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate ∫x2exdx using integration by parts, we will use the formula
∫u dv =uv −∫v du where uand dv are functions of x.
Step 1: Let u=x2and dv =exdx. Then, we have:
du = 2x dx and v=∫exdx =ex
Step 2: Applying the integration by parts formula, we have:
∫x2exdx =x2ex−∫2xexdx
=x2ex−2∫xexdx
Step 3: Let’s now evaluate the remaining integral ∫xexdx. We again use
integration by parts with u=xand dv =exdx:
du =dx and v=∫exdx =ex
Step 4: Applying the integration by parts formula to the integral ∫xexdx,
we get:
∫xexdx =xex−∫exdx
=xex−ex
Step 5: Substituting this back into our original integral, we have:
∫x2exdx =x2ex−2∫xexdx
=x2ex−2(xex−ex)
=x2ex−2xex+ 2ex+C
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C, where Cis the constant of
integration.
8
Question 11
Question
Evaluate the integral ∫xcos(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xcos(x)dx using integration by parts, we will
apply the formula
∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
Step 1: Let’s choose u=xand dv = cos(x)dx. Then, we have:
du =dx
v=∫cos(x)dx = sin(x)
Step 2: Use the integration by parts formula:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C,
where Cis the constant of integration.
Hence, the result of the given integral is ∫xcos(x)dx =xsin(x)+cos(x)+C.
Question 12
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate the given integral ∫x2exdx using integration by parts, we will let
u=x2and dv =exdx. Then, we will find du and vas follows:
•du =d
dx (x2)dx = 2x dx
•v=∫exdx =ex
9
Now, we can apply the integration by parts formula:
∫u dv =uv −∫v du
Step 1: Apply the integration by parts formula.
∫x2exdx =x2ex−∫ex·2x dx
=x2ex−2∫xexdx
Now, we need to evaluate the remaining integral ∫xexdx using integration
by parts again.
Let u=xand dv =exdx. Then, find du and v:
•du =d
dx (x)dx =dx
•v=∫exdx =ex
Step 2: Apply integration by parts to the integral ∫xexdx.
∫xexdx =xex−∫exdx
=xex−ex
Now, substitute this result back into our original integral to solve for the
final answer.
Step 3: Substitute the result back into the original integral.
∫x2exdx =x2ex−2 (xex−ex)
=x2ex−2xex+ 2ex+C
So, ∫x2exdx =x2ex−2xex+2ex+C, where Cis the constant of integration.
Question 13
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To find the integral of x2ex, we will use integration by parts. Let’s choose our
functions uand dv such that du is easy to compute and vis easy to differentiate.
Let u=x2and dv =exdx. Then, we have:
10
] Compute du and v.
du = 2x dx
To find v, integrate dv with respect to x.
v=∫exdx =ex] Apply integration by parts formula:
∫u dv =uv −∫v du
This gives us:
∫x2exdx =x2ex−∫2xexdx
] Integrate the remaining integral by parts: Let u=xand dv = 2exdx.
Compute du and vas follows:
du =dx
v=∫2exdx = 2ex
] Apply integration by parts formula again:
∫2xexdx = 2xex−∫2exdx
Substitute back into the previous equation:
∫x2exdx =x2ex−(2xex−2ex)
∫x2exdx =x2ex−2xex+ 2ex+C
where Cis the constant of integration.
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C.
Question 14
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will apply
the formula: ∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
11
Step 1: Let’s choose u= ln(x)and dv =x2dx. Then, du =1
xdx and
v=1
3x3.
Step 2: Now, let’s apply the integration by parts formula:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C,
where Cis the constant of integration.
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C.
Question 15
Question
Evaluate the definite integral ∫1
0xsin−1(x)dx using integration by parts.
Solution
To evaluate the given definite integral ∫1
0xsin−1(x)dx using integration by
parts, we will use the formula ∫u dv =uv −∫v du.
Step 1: Choose uand dv
Let u= sin−1(x)and dv =x dx. Then, du =1
√1−x2dx and v=1
2x2.
Step 2: Apply integration by parts
∫xsin−1(x)dx =uv −∫v du
= sin−1(x)(1
2x2)−∫(1
2x2·1
√1−x2)dx
=1
2x2sin−1(x)−1
2∫x2
√1−x2dx
Step 3: Evaluate the remaining integral
To evaluate ∫x2
√1−x2dx, we can use the substitution u= 1−x2,du =−2x dx:
∫x2
√1−x2dx =−1
2∫−2x
√1−x2·x dx
=−1
2∫−du
√u
=√u+C
=√1−x2+C
12
Step 4: Substitute back and apply limits
∫1
0
xsin−1(x)dx =[1
2x2sin−1(x)−1
2(√1−x2)]1
0
=(1
2sin−1(1) −1
2√1−1)−(1
2sin−1(0) −1
2√1−0)
=π
4
Therefore, ∫1
0xsin−1(x)dx =π
4.
Question 16
Question
Evaluate the integral: ∫x2exdx.
Solution
To evaluate the given integral, we will use integration by parts. Integration by
parts is given by the formula:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let u=x2and dv =exdx. Then, calculate du and v:
du = 2x dx and v=∫exdx =ex
Step 2: Apply the formula for integration by parts:
∫x2exdx =x2ex−∫2xexdx
Step 3: Now, we need to evaluate the remaining integral on the right-hand
side, ∫2xexdx, using integration by parts again.
Step 4: Let u= 2xand dv =exdx. Calculate du and v:
du = 2 dx and v=ex
Step 5: Apply the formula for integration by parts to evaluate ∫2xexdx:
∫2xexdx = 2xex−∫2exdx
13
Step 6: Simplify the integral:
∫2exdx = 2 ∫exdx = 2ex
Step 7: Substitute back into our original integral:
∫x2exdx =x2ex−2xex+C
Therefore, ∫x2exdx =x2ex−2xex+C, where Cis the constant of integra-
tion.
Question 17
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx, we can use the integration by parts for-
mula: ∫u dv =uv −∫v du.
Let u= ln(x)and dv =x dx. Then, we have:
du =1
xdx
v=1
2x2
14
Now, we will apply the integration by parts formula:
Step 1: Let u= ln(x), dv =x dx
Step 2: Calculate du and v:
du =1
xdx, v =1
2x2
Step 3: Apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)(1
2x2)−∫(1
2x2)(1
x)dx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 18
Question
Compute the integral ∫x2exdx using integration by parts.
Solution
To evaluate the given integral ∫x2exdx using integration by parts, we will apply
the formula ∫u dv =uv −∫v du with u=x2and dv =exdx.
Step 1: Let’s determine du and v. We have:
du =d
dx (x2)dx = 2x dx
v=∫exdx =ex
Step 2: Apply integration by parts formula.
∫x2exdx =x2ex−∫ex·2x dx
=x2ex−2∫xexdx
15
Now, we need to apply integration by parts again to evaluate the remaining
integral ∫xexdx.
Step 3: Let’s apply integration by parts to ∫xexdx. Choose u=xand
dv =exdx.
Now we determine du and v:
du =dx
v=∫exdx =ex
Step 4: Apply integration by parts formula to ∫xexdx.
∫xexdx =xex−∫exdx
=xex−ex
Step 5: Substitute back into our original integral.
∫x2exdx =x2ex−2∫xexdx
=x2ex−2(xex−ex)
=x2ex−2xex+ 2ex+C
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C, where Cis the constant of
integration.
Question 19
Question
Evaluate the following integral:
∫x2cos x dx
Solution
To evaluate the given integral ∫x2cos x dx, we will use integration by parts.
Step 1: Let’s choose our uand dv. Let u=x2and dv = cos x dx.
Step 2: Compute du and v.
•du =d
dx (x2)dx = 2x dx
•v=∫cos x dx = sin x
16
Step 3: Apply the formula for integration by parts.
∫u dv =uv −∫v du
Now, we substitute u,dv,du, and vinto the formula:
∫x2cos x dx =x2sin x−∫sin x·2x dx
Step 4: Simplify the resulting integral.
∫x2cos x dx =x2sin x−2∫xsin x dx
Step 5: Let’s use integration by parts again for the remaining integral. Let
u=xand dv = sin x dx.
Now, compute du and v:
•du =d
dx (x)dx =dx
•v=∫sin x dx =−cos x
Step 6: Apply the formula for integration by parts again.
∫x2cos x dx =x2sin x−2(x(−cos x)−∫(−cos x)dx)
Simplify further to get the final result:
∫x2cos x dx =x2sin x+ 2xcos x−2 sin x+C
Where Cis the constant of integration.
Question 20
Question
Evaluate the integral ∫x2sin−1x dx using integration by parts.
Solution
To evaluate the given integral ∫x2sin−1x dx using integration by parts, we need
to choose two functions to assign as uand dv in the formula:
∫u dv =uv −∫v du
Let’s choose u= sin−1xand dv =x2dx. Then, we can find du and v:
du =1
√1−x2dx
17
v=1
3x3
Now, we can apply the integration by parts formula:
∫x2sin−1x dx =1
3x3sin−1x−∫1
3x3·1
√1−x2dx
∫x2sin−1x dx =1
3x3sin−1x−1
3∫x3
√1−x2dx
Integrating the remaining integral ∫x3
√1−x2dx would require further manip-
ulation using substitution or other techniques.
Question 21
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x2ln(x)dx, we will use integration by parts, which
states ∫u dv =uv −∫v du.
Let u= ln(x)and dv =x2dx. Then, we have
du =1
xdx and v=1
3x3.
Step 1: Apply integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx.
Step 2: Integrate the remaining integral:
∫x2dx =1
3x3+C,
where Cis the constant of integration.
Step 3: Substitute back to find the final answer:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
=1
3x3(ln(x)−1
3) + C.
Therefore, ∫x2ln(x)dx =1
3x3(ln(x)−1
3) + C.
18
Question 22
Question
Evaluate the integral
∫xe2xdx
Solution
To evaluate the given integral, we will use the technique of integration by parts,
which states ∫u dv =uv −∫v du
Let u=xand dv =e2xdx. Then, we have du =dx and v=1
2e2x.
Step 1: Apply integration by parts:
∫xe2xdx =x(1
2e2x)−∫(1
2e2x)dx
=1
2xe2x−1
4e2x+C
Step 2: Simplify the result:
∫xe2xdx =1
2xe2x−1
4e2x+C
Therefore, the integral ∫xe2xdx evaluates to 1
2xe2x−1
4e2x+C.
Question 23
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will have to
choose which part of the integrand to differentiate and which part to integrate.
Let’s follow the formula for integration by parts:
∫u dv =uv −∫v du
Step 1: Choose u= ln(x)and dv =x dx.
Then, differentiate uto get:
du =1
xdx
19
And integrate dv to get:
v=x2
2
Step 2: Apply the integration by parts formula:
∫xln(x)dx = ln(x)·x2
2−∫x2
2·1
xdx
Simplify the expression:
∫xln(x)dx =x2
2ln(x)−1
2∫x dx
Step 3: Integrate the remaining integral:
∫x dx =x2
2
Step 4: Substitute back into the equation:
∫xln(x)dx =x2
2ln(x)−1
2·x2
2+C
∫xln(x)dx =x2
2ln(x)−x2
4+C
Therefore, the result of the given integral is x2
2ln(x)−x2
4+C, where Cis
the constant of integration.
Question 24
Question
Evaluate the integral ∫excos(x)dx using integration by parts.
Solution
To evaluate the given integral ∫excos(x)dx using integration by parts, we will
apply the formula:
∫u dv =uv −∫v du
Step 1: Let’s choose uand dv:
u=exand dv = cos(x)dx
Step 2: Compute du and v:
du =exdx
20
v=∫cos(x)dx = sin(x)
Step 3: Apply the integration by parts formula:
∫excos(x)dx =exsin(x)−∫sin(x)·exdx
Step 4: Now we have a new integral to evaluate. Let’s again apply integra-
tion by parts with u= sin(x)and dv =exdx:
u= sin(x)and dv =exdx
du = cos(x)dx
v=∫exdx =ex
Step 5: Apply the integration by parts formula:
∫sin(x)·exdx =exsin(x)−∫excos(x)dx
Step 6: Substitute this back into the first equation:
∫excos(x)dx =exsin(x)−(exsin(x)−∫excos(x)dx)
Step 7: Simplify and solve for the original integral:
2∫excos(x)dx =exsin(x)
∫excos(x)dx =exsin(x)
2+C
Therefore, ∫excos(x)dx =exsin(x)
2+C, where Cis the constant of integra-
tion.
Question 25
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
21
Solution
To integrate ∫xln(x)dx, we will use integration by parts where we will set
u= ln(x)and dv =x dx.
Step 1: Compute du and v.
•u= ln(x)
•du =1
xdx
•dv =x dx =⇒v=1
2x2
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C(where Cis the constant of integration)
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C.
22
Step 5: Simplify the integral. Simplify the above expression to get:
1
2x2ln(x)−1
2∫x dx
Step 6: Evaluate the remaining integral. Now, integrate the remaining
term: ∫x dx =1
2x2
Step 7: Final answer. Substitute this back into the expression to get:
1
2x2ln(x)−1
4x2+C
where Cis the constant of integration.
Question 2
Question
Evaluate the integral ∫xcos(x)dx using integration by parts.
Solution
To evaluate the integral ∫xcos(x)dx using integration by parts, we will use the
formula ∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
Step 1: Let’s choose u=xand dv = cos(x)dx. Then, we have:
du =dx and v=∫cos(x)dx = sin(x).
Step 2: Now we can apply the integration by parts formula:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C,
where Cis the constant of integration.
Therefore, the integral ∫xcos(x)dx evaluates to xsin(x) + cos(x) + C.
2
Question 3
Question
Calculate the following integral using integration by parts:
∫x2ln(x)dx
Solution
To evaluate the integral ∫x2ln(x)dx, we’ll use integration by parts. Recall the
formula for integration by parts:
∫u dv =uv −∫v du
where uand vare differentiable functions of x. Let’s choose u= ln(x)and
dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Step 1: Calculate du and v:
du =1
xdx and v=1
3x3
Step 2: Apply the formula for integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 3: Evaluate the last integral:
∫x2dx =1
3x3+C
Step 4: Substitute the result back into the formula:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
Therefore, the integral ∫x2ln(x)dx simplifies to 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
Question 4
Question
Evaluate the integral ∫xexdx using integration by parts.
3
Solution
To evaluate the integral ∫xexdx using integration by parts, we will apply the
formula ∫u dv =uv −∫v du with u=xand dv =exdx.
Step 1: Choose uand dv. Let u=xand dv =exdx.
Step 2: Compute du and v. Compute du by taking the derivative of uwith
respect to x:du =dx
Integrate dv to find v:v=∫exdx =ex
Step 3: Apply the integration by parts formula. We have: ∫xexdx =
uv −∫v du
=x·ex−∫exdx
Step 4: Evaluate the integral. Continuing from where we left off: =x·ex−
∫exdx
=x·ex−ex+C
Therefore, ∫xexdx =x·ex−ex+C, where Cis the constant of integration.
Question 5
Question
Evaluate the integral ∫xln(x)dx by using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let u= ln(x)and dv =x dx. Then, du =1
xdx and v=1
2x2.
Step 2: Now, we can apply the integration by parts formula:
∫xln(x)dx =∫u dv
=uv −∫v du
= ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
4
Question 6
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To solve this integral, we will use integration by parts, which states that ∫u dv =
uv −∫v du. In this case, we will let u= ln(x)and dv =x2dx.
Step 1: Find du and v.
•u= ln(x)
•du =1
xdx
•v=∫x2dx =x3
3
Step 2: Apply integration by parts formula.
∫x2ln(x)dx =uv −∫v du
= ln(x)·x3
3−∫x3
3·1
xdx
=x3ln(x)
3−1
3∫x2dx
Step 3: Integrate the remaining integral.
∫x2ln(x)dx =x3ln(x)
3−1
3∫x2dx
=x3ln(x)
3−1
3·x3
3+C
=x3ln(x)
3−x3
9+C
Therefore, the solution to the integral ∫x2ln(x)dx is x3ln(x)
3−x3
9+C, where
Cis the constant of integration.
Question 7
Question
Evaluate the integral ∫xexdx using integration by parts.
5
Solution
To evaluate the integral ∫xexdx using integration by parts, we will apply the
formula ∫u dv =uv −∫v du.
Step 1: We will choose uand dv. Let u=xand dv =exdx.
Step 2: Calculate du and v.
•du =dx
• To find v, we integrate dv:
∫exdx =ex+C
So, v=ex
Step 3: Apply the integration by parts formula.
∫xexdx =uv −∫v du
=x·ex−∫exdx
Step 4: Simplify the integral.
=x·ex−ex+C= (x−1)ex+C
Thus, ∫xexdx = (x−1)ex+C.
Question 8
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫x2ln(x)dx using integration by parts, we will
choose u= ln(x)and dv =x2dx.
Step 1: Calculate du and v.
•u= ln(x)
•du =1
xdx
•v=∫x2dx =1
3x3
6
Step 2: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C, where Cis the constant of
integration.
Question 9
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate ∫xln(x)dx, we will use integration by parts. Recall the formula
for integration by parts: ∫u dv =uv −∫v du.
Step 1: Let us choose: u= ln(x)
dv =x dx
Differentiating uand integrating dv:du =1
xdx
v=1
2x2
Step 2: Write out the formula ∫u dv =uv −∫v du:∫xln(x)dx =
ln(x)·1
2x2−∫1
2x2·1
xdx
Simplify: =1
2x2ln(x)−1
2∫x dx
Step 3: Now, we integrate ∫x dx:=1
2x2ln(x)−1
2·1
2x2+C
Finally, simplifying the expression gives: =1
2x2ln(x)−1
4x2+C, where C
is the constant of integration.
7
Question 10
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate ∫x2exdx using integration by parts, we will use the formula
∫u dv =uv −∫v du where uand dv are functions of x.
Step 1: Let u=x2and dv =exdx. Then, we have:
du = 2x dx and v=∫exdx =ex
Step 2: Applying the integration by parts formula, we have:
∫x2exdx =x2ex−∫2xexdx
=x2ex−2∫xexdx
Step 3: Let’s now evaluate the remaining integral ∫xexdx. We again use
integration by parts with u=xand dv =exdx:
du =dx and v=∫exdx =ex
Step 4: Applying the integration by parts formula to the integral ∫xexdx,
we get:
∫xexdx =xex−∫exdx
=xex−ex
Step 5: Substituting this back into our original integral, we have:
∫x2exdx =x2ex−2∫xexdx
=x2ex−2(xex−ex)
=x2ex−2xex+ 2ex+C
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C, where Cis the constant of
integration.
8
Question 11
Question
Evaluate the integral ∫xcos(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xcos(x)dx using integration by parts, we will
apply the formula
∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
Step 1: Let’s choose u=xand dv = cos(x)dx. Then, we have:
du =dx
v=∫cos(x)dx = sin(x)
Step 2: Use the integration by parts formula:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C,
where Cis the constant of integration.
Hence, the result of the given integral is ∫xcos(x)dx =xsin(x)+cos(x)+C.
Question 12
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate the given integral ∫x2exdx using integration by parts, we will let
u=x2and dv =exdx. Then, we will find du and vas follows:
•du =d
dx (x2)dx = 2x dx
•v=∫exdx =ex
9
Now, we can apply the integration by parts formula:
∫u dv =uv −∫v du
Step 1: Apply the integration by parts formula.
∫x2exdx =x2ex−∫ex·2x dx
=x2ex−2∫xexdx
Now, we need to evaluate the remaining integral ∫xexdx using integration
by parts again.
Let u=xand dv =exdx. Then, find du and v:
•du =d
dx (x)dx =dx
•v=∫exdx =ex
Step 2: Apply integration by parts to the integral ∫xexdx.
∫xexdx =xex−∫exdx
=xex−ex
Now, substitute this result back into our original integral to solve for the
final answer.
Step 3: Substitute the result back into the original integral.
∫x2exdx =x2ex−2 (xex−ex)
=x2ex−2xex+ 2ex+C
So, ∫x2exdx =x2ex−2xex+2ex+C, where Cis the constant of integration.
Question 13
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To find the integral of x2ex, we will use integration by parts. Let’s choose our
functions uand dv such that du is easy to compute and vis easy to differentiate.
Let u=x2and dv =exdx. Then, we have:
10
] Compute du and v.
du = 2x dx
To find v, integrate dv with respect to x.
v=∫exdx =ex] Apply integration by parts formula:
∫u dv =uv −∫v du
This gives us:
∫x2exdx =x2ex−∫2xexdx
] Integrate the remaining integral by parts: Let u=xand dv = 2exdx.
Compute du and vas follows:
du =dx
v=∫2exdx = 2ex
] Apply integration by parts formula again:
∫2xexdx = 2xex−∫2exdx
Substitute back into the previous equation:
∫x2exdx =x2ex−(2xex−2ex)
∫x2exdx =x2ex−2xex+ 2ex+C
where Cis the constant of integration.
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C.
Question 14
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will apply
the formula: ∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
11
Step 1: Let’s choose u= ln(x)and dv =x2dx. Then, du =1
xdx and
v=1
3x3.
Step 2: Now, let’s apply the integration by parts formula:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C,
where Cis the constant of integration.
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C.
Question 15
Question
Evaluate the definite integral ∫1
0xsin−1(x)dx using integration by parts.
Solution
To evaluate the given definite integral ∫1
0xsin−1(x)dx using integration by
parts, we will use the formula ∫u dv =uv −∫v du.
Step 1: Choose uand dv
Let u= sin−1(x)and dv =x dx. Then, du =1
√1−x2dx and v=1
2x2.
Step 2: Apply integration by parts
∫xsin−1(x)dx =uv −∫v du
= sin−1(x)(1
2x2)−∫(1
2x2·1
√1−x2)dx
=1
2x2sin−1(x)−1
2∫x2
√1−x2dx
Step 3: Evaluate the remaining integral
To evaluate ∫x2
√1−x2dx, we can use the substitution u= 1−x2,du =−2x dx:
∫x2
√1−x2dx =−1
2∫−2x
√1−x2·x dx
=−1
2∫−du
√u
=√u+C
=√1−x2+C
12
Step 4: Substitute back and apply limits
∫1
0
xsin−1(x)dx =[1
2x2sin−1(x)−1
2(√1−x2)]1
0
=(1
2sin−1(1) −1
2√1−1)−(1
2sin−1(0) −1
2√1−0)
=π
4
Therefore, ∫1
0xsin−1(x)dx =π
4.
Question 16
Question
Evaluate the integral: ∫x2exdx.
Solution
To evaluate the given integral, we will use integration by parts. Integration by
parts is given by the formula:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let u=x2and dv =exdx. Then, calculate du and v:
du = 2x dx and v=∫exdx =ex
Step 2: Apply the formula for integration by parts:
∫x2exdx =x2ex−∫2xexdx
Step 3: Now, we need to evaluate the remaining integral on the right-hand
side, ∫2xexdx, using integration by parts again.
Step 4: Let u= 2xand dv =exdx. Calculate du and v:
du = 2 dx and v=ex
Step 5: Apply the formula for integration by parts to evaluate ∫2xexdx:
∫2xexdx = 2xex−∫2exdx
13
Step 6: Simplify the integral:
∫2exdx = 2 ∫exdx = 2ex
Step 7: Substitute back into our original integral:
∫x2exdx =x2ex−2xex+C
Therefore, ∫x2exdx =x2ex−2xex+C, where Cis the constant of integra-
tion.
Question 17
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx, we can use the integration by parts for-
mula: ∫u dv =uv −∫v du.
Let u= ln(x)and dv =x dx. Then, we have:
du =1
xdx
v=1
2x2
14
Now, we will apply the integration by parts formula:
Step 1: Let u= ln(x), dv =x dx
Step 2: Calculate du and v:
du =1
xdx, v =1
2x2
Step 3: Apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)(1
2x2)−∫(1
2x2)(1
x)dx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 18
Question
Compute the integral ∫x2exdx using integration by parts.
Solution
To evaluate the given integral ∫x2exdx using integration by parts, we will apply
the formula ∫u dv =uv −∫v du with u=x2and dv =exdx.
Step 1: Let’s determine du and v. We have:
du =d
dx (x2)dx = 2x dx
v=∫exdx =ex
Step 2: Apply integration by parts formula.
∫x2exdx =x2ex−∫ex·2x dx
=x2ex−2∫xexdx
15
Now, we need to apply integration by parts again to evaluate the remaining
integral ∫xexdx.
Step 3: Let’s apply integration by parts to ∫xexdx. Choose u=xand
dv =exdx.
Now we determine du and v:
du =dx
v=∫exdx =ex
Step 4: Apply integration by parts formula to ∫xexdx.
∫xexdx =xex−∫exdx
=xex−ex
Step 5: Substitute back into our original integral.
∫x2exdx =x2ex−2∫xexdx
=x2ex−2(xex−ex)
=x2ex−2xex+ 2ex+C
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C, where Cis the constant of
integration.
Question 19
Question
Evaluate the following integral:
∫x2cos x dx
Solution
To evaluate the given integral ∫x2cos x dx, we will use integration by parts.
Step 1: Let’s choose our uand dv. Let u=x2and dv = cos x dx.
Step 2: Compute du and v.
•du =d
dx (x2)dx = 2x dx
•v=∫cos x dx = sin x
16
Step 3: Apply the formula for integration by parts.
∫u dv =uv −∫v du
Now, we substitute u,dv,du, and vinto the formula:
∫x2cos x dx =x2sin x−∫sin x·2x dx
Step 4: Simplify the resulting integral.
∫x2cos x dx =x2sin x−2∫xsin x dx
Step 5: Let’s use integration by parts again for the remaining integral. Let
u=xand dv = sin x dx.
Now, compute du and v:
•du =d
dx (x)dx =dx
•v=∫sin x dx =−cos x
Step 6: Apply the formula for integration by parts again.
∫x2cos x dx =x2sin x−2(x(−cos x)−∫(−cos x)dx)
Simplify further to get the final result:
∫x2cos x dx =x2sin x+ 2xcos x−2 sin x+C
Where Cis the constant of integration.
Question 20
Question
Evaluate the integral ∫x2sin−1x dx using integration by parts.
Solution
To evaluate the given integral ∫x2sin−1x dx using integration by parts, we need
to choose two functions to assign as uand dv in the formula:
∫u dv =uv −∫v du
Let’s choose u= sin−1xand dv =x2dx. Then, we can find du and v:
du =1
√1−x2dx
17
v=1
3x3
Now, we can apply the integration by parts formula:
∫x2sin−1x dx =1
3x3sin−1x−∫1
3x3·1
√1−x2dx
∫x2sin−1x dx =1
3x3sin−1x−1
3∫x3
√1−x2dx
Integrating the remaining integral ∫x3
√1−x2dx would require further manip-
ulation using substitution or other techniques.
Question 21
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x2ln(x)dx, we will use integration by parts, which
states ∫u dv =uv −∫v du.
Let u= ln(x)and dv =x2dx. Then, we have
du =1
xdx and v=1
3x3.
Step 1: Apply integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx.
Step 2: Integrate the remaining integral:
∫x2dx =1
3x3+C,
where Cis the constant of integration.
Step 3: Substitute back to find the final answer:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
=1
3x3(ln(x)−1
3) + C.
Therefore, ∫x2ln(x)dx =1
3x3(ln(x)−1
3) + C.
18
Question 22
Question
Evaluate the integral
∫xe2xdx
Solution
To evaluate the given integral, we will use the technique of integration by parts,
which states ∫u dv =uv −∫v du
Let u=xand dv =e2xdx. Then, we have du =dx and v=1
2e2x.
Step 1: Apply integration by parts:
∫xe2xdx =x(1
2e2x)−∫(1
2e2x)dx
=1
2xe2x−1
4e2x+C
Step 2: Simplify the result:
∫xe2xdx =1
2xe2x−1
4e2x+C
Therefore, the integral ∫xe2xdx evaluates to 1
2xe2x−1
4e2x+C.
Question 23
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will have to
choose which part of the integrand to differentiate and which part to integrate.
Let’s follow the formula for integration by parts:
∫u dv =uv −∫v du
Step 1: Choose u= ln(x)and dv =x dx.
Then, differentiate uto get:
du =1
xdx
19
And integrate dv to get:
v=x2
2
Step 2: Apply the integration by parts formula:
∫xln(x)dx = ln(x)·x2
2−∫x2
2·1
xdx
Simplify the expression:
∫xln(x)dx =x2
2ln(x)−1
2∫x dx
Step 3: Integrate the remaining integral:
∫x dx =x2
2
Step 4: Substitute back into the equation:
∫xln(x)dx =x2
2ln(x)−1
2·x2
2+C
∫xln(x)dx =x2
2ln(x)−x2
4+C
Therefore, the result of the given integral is x2
2ln(x)−x2
4+C, where Cis
the constant of integration.
Question 24
Question
Evaluate the integral ∫excos(x)dx using integration by parts.
Solution
To evaluate the given integral ∫excos(x)dx using integration by parts, we will
apply the formula:
∫u dv =uv −∫v du
Step 1: Let’s choose uand dv:
u=exand dv = cos(x)dx
Step 2: Compute du and v:
du =exdx
20
v=∫cos(x)dx = sin(x)
Step 3: Apply the integration by parts formula:
∫excos(x)dx =exsin(x)−∫sin(x)·exdx
Step 4: Now we have a new integral to evaluate. Let’s again apply integra-
tion by parts with u= sin(x)and dv =exdx:
u= sin(x)and dv =exdx
du = cos(x)dx
v=∫exdx =ex
Step 5: Apply the integration by parts formula:
∫sin(x)·exdx =exsin(x)−∫excos(x)dx
Step 6: Substitute this back into the first equation:
∫excos(x)dx =exsin(x)−(exsin(x)−∫excos(x)dx)
Step 7: Simplify and solve for the original integral:
2∫excos(x)dx =exsin(x)
∫excos(x)dx =exsin(x)
2+C
Therefore, ∫excos(x)dx =exsin(x)
2+C, where Cis the constant of integra-
tion.
Question 25
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
21
Solution
To integrate ∫xln(x)dx, we will use integration by parts where we will set
u= ln(x)and dv =x dx.
Step 1: Compute du and v.
•u= ln(x)
•du =1
xdx
•dv =x dx =⇒v=1
2x2
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C(where Cis the constant of integration)
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C.
22
Step 5: Simplify the integral. Simplify the above expression to get:
1
2x2ln(x)−1
2∫x dx
Step 6: Evaluate the remaining integral. Now, integrate the remaining
term: ∫x dx =1
2x2
Step 7: Final answer. Substitute this back into the expression to get:
1
2x2ln(x)−1
4x2+C
where Cis the constant of integration.
Question 2
Question
Evaluate the integral ∫xcos(x)dx using integration by parts.
Solution
To evaluate the integral ∫xcos(x)dx using integration by parts, we will use the
formula ∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
Step 1: Let’s choose u=xand dv = cos(x)dx. Then, we have:
du =dx and v=∫cos(x)dx = sin(x).
Step 2: Now we can apply the integration by parts formula:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C,
where Cis the constant of integration.
Therefore, the integral ∫xcos(x)dx evaluates to xsin(x) + cos(x) + C.
2
Question 3
Question
Calculate the following integral using integration by parts:
∫x2ln(x)dx
Solution
To evaluate the integral ∫x2ln(x)dx, we’ll use integration by parts. Recall the
formula for integration by parts:
∫u dv =uv −∫v du
where uand vare differentiable functions of x. Let’s choose u= ln(x)and
dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Step 1: Calculate du and v:
du =1
xdx and v=1
3x3
Step 2: Apply the formula for integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 3: Evaluate the last integral:
∫x2dx =1
3x3+C
Step 4: Substitute the result back into the formula:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
Therefore, the integral ∫x2ln(x)dx simplifies to 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
Question 4
Question
Evaluate the integral ∫xexdx using integration by parts.
3
Solution
To evaluate the integral ∫xexdx using integration by parts, we will apply the
formula ∫u dv =uv −∫v du with u=xand dv =exdx.
Step 1: Choose uand dv. Let u=xand dv =exdx.
Step 2: Compute du and v. Compute du by taking the derivative of uwith
respect to x:du =dx
Integrate dv to find v:v=∫exdx =ex
Step 3: Apply the integration by parts formula. We have: ∫xexdx =
uv −∫v du
=x·ex−∫exdx
Step 4: Evaluate the integral. Continuing from where we left off: =x·ex−
∫exdx
=x·ex−ex+C
Therefore, ∫xexdx =x·ex−ex+C, where Cis the constant of integration.
Question 5
Question
Evaluate the integral ∫xln(x)dx by using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let u= ln(x)and dv =x dx. Then, du =1
xdx and v=1
2x2.
Step 2: Now, we can apply the integration by parts formula:
∫xln(x)dx =∫u dv
=uv −∫v du
= ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
4
Question 6
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To solve this integral, we will use integration by parts, which states that ∫u dv =
uv −∫v du. In this case, we will let u= ln(x)and dv =x2dx.
Step 1: Find du and v.
•u= ln(x)
•du =1
xdx
•v=∫x2dx =x3
3
Step 2: Apply integration by parts formula.
∫x2ln(x)dx =uv −∫v du
= ln(x)·x3
3−∫x3
3·1
xdx
=x3ln(x)
3−1
3∫x2dx
Step 3: Integrate the remaining integral.
∫x2ln(x)dx =x3ln(x)
3−1
3∫x2dx
=x3ln(x)
3−1
3·x3
3+C
=x3ln(x)
3−x3
9+C
Therefore, the solution to the integral ∫x2ln(x)dx is x3ln(x)
3−x3
9+C, where
Cis the constant of integration.
Question 7
Question
Evaluate the integral ∫xexdx using integration by parts.
5
Solution
To evaluate the integral ∫xexdx using integration by parts, we will apply the
formula ∫u dv =uv −∫v du.
Step 1: We will choose uand dv. Let u=xand dv =exdx.
Step 2: Calculate du and v.
•du =dx
• To find v, we integrate dv:
∫exdx =ex+C
So, v=ex
Step 3: Apply the integration by parts formula.
∫xexdx =uv −∫v du
=x·ex−∫exdx
Step 4: Simplify the integral.
=x·ex−ex+C= (x−1)ex+C
Thus, ∫xexdx = (x−1)ex+C.
Question 8
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫x2ln(x)dx using integration by parts, we will
choose u= ln(x)and dv =x2dx.
Step 1: Calculate du and v.
•u= ln(x)
•du =1
xdx
•v=∫x2dx =1
3x3
6
Step 2: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C, where Cis the constant of
integration.
Question 9
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate ∫xln(x)dx, we will use integration by parts. Recall the formula
for integration by parts: ∫u dv =uv −∫v du.
Step 1: Let us choose: u= ln(x)
dv =x dx
Differentiating uand integrating dv:du =1
xdx
v=1
2x2
Step 2: Write out the formula ∫u dv =uv −∫v du:∫xln(x)dx =
ln(x)·1
2x2−∫1
2x2·1
xdx
Simplify: =1
2x2ln(x)−1
2∫x dx
Step 3: Now, we integrate ∫x dx:=1
2x2ln(x)−1
2·1
2x2+C
Finally, simplifying the expression gives: =1
2x2ln(x)−1
4x2+C, where C
is the constant of integration.
7
Question 10
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate ∫x2exdx using integration by parts, we will use the formula
∫u dv =uv −∫v du where uand dv are functions of x.
Step 1: Let u=x2and dv =exdx. Then, we have:
du = 2x dx and v=∫exdx =ex
Step 2: Applying the integration by parts formula, we have:
∫x2exdx =x2ex−∫2xexdx
=x2ex−2∫xexdx
Step 3: Let’s now evaluate the remaining integral ∫xexdx. We again use
integration by parts with u=xand dv =exdx:
du =dx and v=∫exdx =ex
Step 4: Applying the integration by parts formula to the integral ∫xexdx,
we get:
∫xexdx =xex−∫exdx
=xex−ex
Step 5: Substituting this back into our original integral, we have:
∫x2exdx =x2ex−2∫xexdx
=x2ex−2(xex−ex)
=x2ex−2xex+ 2ex+C
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C, where Cis the constant of
integration.
8
Question 11
Question
Evaluate the integral ∫xcos(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xcos(x)dx using integration by parts, we will
apply the formula
∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
Step 1: Let’s choose u=xand dv = cos(x)dx. Then, we have:
du =dx
v=∫cos(x)dx = sin(x)
Step 2: Use the integration by parts formula:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C,
where Cis the constant of integration.
Hence, the result of the given integral is ∫xcos(x)dx =xsin(x)+cos(x)+C.
Question 12
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate the given integral ∫x2exdx using integration by parts, we will let
u=x2and dv =exdx. Then, we will find du and vas follows:
•du =d
dx (x2)dx = 2x dx
•v=∫exdx =ex
9
Now, we can apply the integration by parts formula:
∫u dv =uv −∫v du
Step 1: Apply the integration by parts formula.
∫x2exdx =x2ex−∫ex·2x dx
=x2ex−2∫xexdx
Now, we need to evaluate the remaining integral ∫xexdx using integration
by parts again.
Let u=xand dv =exdx. Then, find du and v:
•du =d
dx (x)dx =dx
•v=∫exdx =ex
Step 2: Apply integration by parts to the integral ∫xexdx.
∫xexdx =xex−∫exdx
=xex−ex
Now, substitute this result back into our original integral to solve for the
final answer.
Step 3: Substitute the result back into the original integral.
∫x2exdx =x2ex−2 (xex−ex)
=x2ex−2xex+ 2ex+C
So, ∫x2exdx =x2ex−2xex+2ex+C, where Cis the constant of integration.
Question 13
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To find the integral of x2ex, we will use integration by parts. Let’s choose our
functions uand dv such that du is easy to compute and vis easy to differentiate.
Let u=x2and dv =exdx. Then, we have:
10
] Compute du and v.
du = 2x dx
To find v, integrate dv with respect to x.
v=∫exdx =ex] Apply integration by parts formula:
∫u dv =uv −∫v du
This gives us:
∫x2exdx =x2ex−∫2xexdx
] Integrate the remaining integral by parts: Let u=xand dv = 2exdx.
Compute du and vas follows:
du =dx
v=∫2exdx = 2ex
] Apply integration by parts formula again:
∫2xexdx = 2xex−∫2exdx
Substitute back into the previous equation:
∫x2exdx =x2ex−(2xex−2ex)
∫x2exdx =x2ex−2xex+ 2ex+C
where Cis the constant of integration.
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C.
Question 14
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will apply
the formula: ∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
11
Step 1: Let’s choose u= ln(x)and dv =x2dx. Then, du =1
xdx and
v=1
3x3.
Step 2: Now, let’s apply the integration by parts formula:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C,
where Cis the constant of integration.
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C.
Question 15
Question
Evaluate the definite integral ∫1
0xsin−1(x)dx using integration by parts.
Solution
To evaluate the given definite integral ∫1
0xsin−1(x)dx using integration by
parts, we will use the formula ∫u dv =uv −∫v du.
Step 1: Choose uand dv
Let u= sin−1(x)and dv =x dx. Then, du =1
√1−x2dx and v=1
2x2.
Step 2: Apply integration by parts
∫xsin−1(x)dx =uv −∫v du
= sin−1(x)(1
2x2)−∫(1
2x2·1
√1−x2)dx
=1
2x2sin−1(x)−1
2∫x2
√1−x2dx
Step 3: Evaluate the remaining integral
To evaluate ∫x2
√1−x2dx, we can use the substitution u= 1−x2,du =−2x dx:
∫x2
√1−x2dx =−1
2∫−2x
√1−x2·x dx
=−1
2∫−du
√u
=√u+C
=√1−x2+C
12
Step 4: Substitute back and apply limits
∫1
0
xsin−1(x)dx =[1
2x2sin−1(x)−1
2(√1−x2)]1
0
=(1
2sin−1(1) −1
2√1−1)−(1
2sin−1(0) −1
2√1−0)
=π
4
Therefore, ∫1
0xsin−1(x)dx =π
4.
Question 16
Question
Evaluate the integral: ∫x2exdx.
Solution
To evaluate the given integral, we will use integration by parts. Integration by
parts is given by the formula:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let u=x2and dv =exdx. Then, calculate du and v:
du = 2x dx and v=∫exdx =ex
Step 2: Apply the formula for integration by parts:
∫x2exdx =x2ex−∫2xexdx
Step 3: Now, we need to evaluate the remaining integral on the right-hand
side, ∫2xexdx, using integration by parts again.
Step 4: Let u= 2xand dv =exdx. Calculate du and v:
du = 2 dx and v=ex
Step 5: Apply the formula for integration by parts to evaluate ∫2xexdx:
∫2xexdx = 2xex−∫2exdx
13
Step 6: Simplify the integral:
∫2exdx = 2 ∫exdx = 2ex
Step 7: Substitute back into our original integral:
∫x2exdx =x2ex−2xex+C
Therefore, ∫x2exdx =x2ex−2xex+C, where Cis the constant of integra-
tion.
Question 17
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx, we can use the integration by parts for-
mula: ∫u dv =uv −∫v du.
Let u= ln(x)and dv =x dx. Then, we have:
du =1
xdx
v=1
2x2
14
Now, we will apply the integration by parts formula:
Step 1: Let u= ln(x), dv =x dx
Step 2: Calculate du and v:
du =1
xdx, v =1
2x2
Step 3: Apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)(1
2x2)−∫(1
2x2)(1
x)dx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 18
Question
Compute the integral ∫x2exdx using integration by parts.
Solution
To evaluate the given integral ∫x2exdx using integration by parts, we will apply
the formula ∫u dv =uv −∫v du with u=x2and dv =exdx.
Step 1: Let’s determine du and v. We have:
du =d
dx (x2)dx = 2x dx
v=∫exdx =ex
Step 2: Apply integration by parts formula.
∫x2exdx =x2ex−∫ex·2x dx
=x2ex−2∫xexdx
15
Now, we need to apply integration by parts again to evaluate the remaining
integral ∫xexdx.
Step 3: Let’s apply integration by parts to ∫xexdx. Choose u=xand
dv =exdx.
Now we determine du and v:
du =dx
v=∫exdx =ex
Step 4: Apply integration by parts formula to ∫xexdx.
∫xexdx =xex−∫exdx
=xex−ex
Step 5: Substitute back into our original integral.
∫x2exdx =x2ex−2∫xexdx
=x2ex−2(xex−ex)
=x2ex−2xex+ 2ex+C
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C, where Cis the constant of
integration.
Question 19
Question
Evaluate the following integral:
∫x2cos x dx
Solution
To evaluate the given integral ∫x2cos x dx, we will use integration by parts.
Step 1: Let’s choose our uand dv. Let u=x2and dv = cos x dx.
Step 2: Compute du and v.
•du =d
dx (x2)dx = 2x dx
•v=∫cos x dx = sin x
16
Step 3: Apply the formula for integration by parts.
∫u dv =uv −∫v du
Now, we substitute u,dv,du, and vinto the formula:
∫x2cos x dx =x2sin x−∫sin x·2x dx
Step 4: Simplify the resulting integral.
∫x2cos x dx =x2sin x−2∫xsin x dx
Step 5: Let’s use integration by parts again for the remaining integral. Let
u=xand dv = sin x dx.
Now, compute du and v:
•du =d
dx (x)dx =dx
•v=∫sin x dx =−cos x
Step 6: Apply the formula for integration by parts again.
∫x2cos x dx =x2sin x−2(x(−cos x)−∫(−cos x)dx)
Simplify further to get the final result:
∫x2cos x dx =x2sin x+ 2xcos x−2 sin x+C
Where Cis the constant of integration.
Question 20
Question
Evaluate the integral ∫x2sin−1x dx using integration by parts.
Solution
To evaluate the given integral ∫x2sin−1x dx using integration by parts, we need
to choose two functions to assign as uand dv in the formula:
∫u dv =uv −∫v du
Let’s choose u= sin−1xand dv =x2dx. Then, we can find du and v:
du =1
√1−x2dx
17
v=1
3x3
Now, we can apply the integration by parts formula:
∫x2sin−1x dx =1
3x3sin−1x−∫1
3x3·1
√1−x2dx
∫x2sin−1x dx =1
3x3sin−1x−1
3∫x3
√1−x2dx
Integrating the remaining integral ∫x3
√1−x2dx would require further manip-
ulation using substitution or other techniques.
Question 21
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x2ln(x)dx, we will use integration by parts, which
states ∫u dv =uv −∫v du.
Let u= ln(x)and dv =x2dx. Then, we have
du =1
xdx and v=1
3x3.
Step 1: Apply integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx.
Step 2: Integrate the remaining integral:
∫x2dx =1
3x3+C,
where Cis the constant of integration.
Step 3: Substitute back to find the final answer:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
=1
3x3(ln(x)−1
3) + C.
Therefore, ∫x2ln(x)dx =1
3x3(ln(x)−1
3) + C.
18
Question 22
Question
Evaluate the integral
∫xe2xdx
Solution
To evaluate the given integral, we will use the technique of integration by parts,
which states ∫u dv =uv −∫v du
Let u=xand dv =e2xdx. Then, we have du =dx and v=1
2e2x.
Step 1: Apply integration by parts:
∫xe2xdx =x(1
2e2x)−∫(1
2e2x)dx
=1
2xe2x−1
4e2x+C
Step 2: Simplify the result:
∫xe2xdx =1
2xe2x−1
4e2x+C
Therefore, the integral ∫xe2xdx evaluates to 1
2xe2x−1
4e2x+C.
Question 23
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will have to
choose which part of the integrand to differentiate and which part to integrate.
Let’s follow the formula for integration by parts:
∫u dv =uv −∫v du
Step 1: Choose u= ln(x)and dv =x dx.
Then, differentiate uto get:
du =1
xdx
19
And integrate dv to get:
v=x2
2
Step 2: Apply the integration by parts formula:
∫xln(x)dx = ln(x)·x2
2−∫x2
2·1
xdx
Simplify the expression:
∫xln(x)dx =x2
2ln(x)−1
2∫x dx
Step 3: Integrate the remaining integral:
∫x dx =x2
2
Step 4: Substitute back into the equation:
∫xln(x)dx =x2
2ln(x)−1
2·x2
2+C
∫xln(x)dx =x2
2ln(x)−x2
4+C
Therefore, the result of the given integral is x2
2ln(x)−x2
4+C, where Cis
the constant of integration.
Question 24
Question
Evaluate the integral ∫excos(x)dx using integration by parts.
Solution
To evaluate the given integral ∫excos(x)dx using integration by parts, we will
apply the formula:
∫u dv =uv −∫v du
Step 1: Let’s choose uand dv:
u=exand dv = cos(x)dx
Step 2: Compute du and v:
du =exdx
20
v=∫cos(x)dx = sin(x)
Step 3: Apply the integration by parts formula:
∫excos(x)dx =exsin(x)−∫sin(x)·exdx
Step 4: Now we have a new integral to evaluate. Let’s again apply integra-
tion by parts with u= sin(x)and dv =exdx:
u= sin(x)and dv =exdx
du = cos(x)dx
v=∫exdx =ex
Step 5: Apply the integration by parts formula:
∫sin(x)·exdx =exsin(x)−∫excos(x)dx
Step 6: Substitute this back into the first equation:
∫excos(x)dx =exsin(x)−(exsin(x)−∫excos(x)dx)
Step 7: Simplify and solve for the original integral:
2∫excos(x)dx =exsin(x)
∫excos(x)dx =exsin(x)
2+C
Therefore, ∫excos(x)dx =exsin(x)
2+C, where Cis the constant of integra-
tion.
Question 25
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
21
Solution
To integrate ∫xln(x)dx, we will use integration by parts where we will set
u= ln(x)and dv =x dx.
Step 1: Compute du and v.
•u= ln(x)
•du =1
xdx
•dv =x dx =⇒v=1
2x2
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C(where Cis the constant of integration)
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C.
22
Step 5: Simplify the integral. Simplify the above expression to get:
1
2x2ln(x)−1
2∫x dx
Step 6: Evaluate the remaining integral. Now, integrate the remaining
term: ∫x dx =1
2x2
Step 7: Final answer. Substitute this back into the expression to get:
1
2x2ln(x)−1
4x2+C
where Cis the constant of integration.
Question 2
Question
Evaluate the integral ∫xcos(x)dx using integration by parts.
Solution
To evaluate the integral ∫xcos(x)dx using integration by parts, we will use the
formula ∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
Step 1: Let’s choose u=xand dv = cos(x)dx. Then, we have:
du =dx and v=∫cos(x)dx = sin(x).
Step 2: Now we can apply the integration by parts formula:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C,
where Cis the constant of integration.
Therefore, the integral ∫xcos(x)dx evaluates to xsin(x) + cos(x) + C.
2
Question 3
Question
Calculate the following integral using integration by parts:
∫x2ln(x)dx
Solution
To evaluate the integral ∫x2ln(x)dx, we’ll use integration by parts. Recall the
formula for integration by parts:
∫u dv =uv −∫v du
where uand vare differentiable functions of x. Let’s choose u= ln(x)and
dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Step 1: Calculate du and v:
du =1
xdx and v=1
3x3
Step 2: Apply the formula for integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 3: Evaluate the last integral:
∫x2dx =1
3x3+C
Step 4: Substitute the result back into the formula:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
Therefore, the integral ∫x2ln(x)dx simplifies to 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
Question 4
Question
Evaluate the integral ∫xexdx using integration by parts.
3
Solution
To evaluate the integral ∫xexdx using integration by parts, we will apply the
formula ∫u dv =uv −∫v du with u=xand dv =exdx.
Step 1: Choose uand dv. Let u=xand dv =exdx.
Step 2: Compute du and v. Compute du by taking the derivative of uwith
respect to x:du =dx
Integrate dv to find v:v=∫exdx =ex
Step 3: Apply the integration by parts formula. We have: ∫xexdx =
uv −∫v du
=x·ex−∫exdx
Step 4: Evaluate the integral. Continuing from where we left off: =x·ex−
∫exdx
=x·ex−ex+C
Therefore, ∫xexdx =x·ex−ex+C, where Cis the constant of integration.
Question 5
Question
Evaluate the integral ∫xln(x)dx by using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let u= ln(x)and dv =x dx. Then, du =1
xdx and v=1
2x2.
Step 2: Now, we can apply the integration by parts formula:
∫xln(x)dx =∫u dv
=uv −∫v du
= ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
4
Question 6
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To solve this integral, we will use integration by parts, which states that ∫u dv =
uv −∫v du. In this case, we will let u= ln(x)and dv =x2dx.
Step 1: Find du and v.
•u= ln(x)
•du =1
xdx
•v=∫x2dx =x3
3
Step 2: Apply integration by parts formula.
∫x2ln(x)dx =uv −∫v du
= ln(x)·x3
3−∫x3
3·1
xdx
=x3ln(x)
3−1
3∫x2dx
Step 3: Integrate the remaining integral.
∫x2ln(x)dx =x3ln(x)
3−1
3∫x2dx
=x3ln(x)
3−1
3·x3
3+C
=x3ln(x)
3−x3
9+C
Therefore, the solution to the integral ∫x2ln(x)dx is x3ln(x)
3−x3
9+C, where
Cis the constant of integration.
Question 7
Question
Evaluate the integral ∫xexdx using integration by parts.
5
Solution
To evaluate the integral ∫xexdx using integration by parts, we will apply the
formula ∫u dv =uv −∫v du.
Step 1: We will choose uand dv. Let u=xand dv =exdx.
Step 2: Calculate du and v.
•du =dx
• To find v, we integrate dv:
∫exdx =ex+C
So, v=ex
Step 3: Apply the integration by parts formula.
∫xexdx =uv −∫v du
=x·ex−∫exdx
Step 4: Simplify the integral.
=x·ex−ex+C= (x−1)ex+C
Thus, ∫xexdx = (x−1)ex+C.
Question 8
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫x2ln(x)dx using integration by parts, we will
choose u= ln(x)and dv =x2dx.
Step 1: Calculate du and v.
•u= ln(x)
•du =1
xdx
•v=∫x2dx =1
3x3
6
Step 2: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C, where Cis the constant of
integration.
Question 9
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate ∫xln(x)dx, we will use integration by parts. Recall the formula
for integration by parts: ∫u dv =uv −∫v du.
Step 1: Let us choose: u= ln(x)
dv =x dx
Differentiating uand integrating dv:du =1
xdx
v=1
2x2
Step 2: Write out the formula ∫u dv =uv −∫v du:∫xln(x)dx =
ln(x)·1
2x2−∫1
2x2·1
xdx
Simplify: =1
2x2ln(x)−1
2∫x dx
Step 3: Now, we integrate ∫x dx:=1
2x2ln(x)−1
2·1
2x2+C
Finally, simplifying the expression gives: =1
2x2ln(x)−1
4x2+C, where C
is the constant of integration.
7
Question 10
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate ∫x2exdx using integration by parts, we will use the formula
∫u dv =uv −∫v du where uand dv are functions of x.
Step 1: Let u=x2and dv =exdx. Then, we have:
du = 2x dx and v=∫exdx =ex
Step 2: Applying the integration by parts formula, we have:
∫x2exdx =x2ex−∫2xexdx
=x2ex−2∫xexdx
Step 3: Let’s now evaluate the remaining integral ∫xexdx. We again use
integration by parts with u=xand dv =exdx:
du =dx and v=∫exdx =ex
Step 4: Applying the integration by parts formula to the integral ∫xexdx,
we get:
∫xexdx =xex−∫exdx
=xex−ex
Step 5: Substituting this back into our original integral, we have:
∫x2exdx =x2ex−2∫xexdx
=x2ex−2(xex−ex)
=x2ex−2xex+ 2ex+C
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C, where Cis the constant of
integration.
8
Question 11
Question
Evaluate the integral ∫xcos(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xcos(x)dx using integration by parts, we will
apply the formula
∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
Step 1: Let’s choose u=xand dv = cos(x)dx. Then, we have:
du =dx
v=∫cos(x)dx = sin(x)
Step 2: Use the integration by parts formula:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C,
where Cis the constant of integration.
Hence, the result of the given integral is ∫xcos(x)dx =xsin(x)+cos(x)+C.
Question 12
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate the given integral ∫x2exdx using integration by parts, we will let
u=x2and dv =exdx. Then, we will find du and vas follows:
•du =d
dx (x2)dx = 2x dx
•v=∫exdx =ex
9
Now, we can apply the integration by parts formula:
∫u dv =uv −∫v du
Step 1: Apply the integration by parts formula.
∫x2exdx =x2ex−∫ex·2x dx
=x2ex−2∫xexdx
Now, we need to evaluate the remaining integral ∫xexdx using integration
by parts again.
Let u=xand dv =exdx. Then, find du and v:
•du =d
dx (x)dx =dx
•v=∫exdx =ex
Step 2: Apply integration by parts to the integral ∫xexdx.
∫xexdx =xex−∫exdx
=xex−ex
Now, substitute this result back into our original integral to solve for the
final answer.
Step 3: Substitute the result back into the original integral.
∫x2exdx =x2ex−2 (xex−ex)
=x2ex−2xex+ 2ex+C
So, ∫x2exdx =x2ex−2xex+2ex+C, where Cis the constant of integration.
Question 13
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To find the integral of x2ex, we will use integration by parts. Let’s choose our
functions uand dv such that du is easy to compute and vis easy to differentiate.
Let u=x2and dv =exdx. Then, we have:
10
] Compute du and v.
du = 2x dx
To find v, integrate dv with respect to x.
v=∫exdx =ex] Apply integration by parts formula:
∫u dv =uv −∫v du
This gives us:
∫x2exdx =x2ex−∫2xexdx
] Integrate the remaining integral by parts: Let u=xand dv = 2exdx.
Compute du and vas follows:
du =dx
v=∫2exdx = 2ex
] Apply integration by parts formula again:
∫2xexdx = 2xex−∫2exdx
Substitute back into the previous equation:
∫x2exdx =x2ex−(2xex−2ex)
∫x2exdx =x2ex−2xex+ 2ex+C
where Cis the constant of integration.
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C.
Question 14
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will apply
the formula: ∫u dv =uv −∫v du,
where uand vare differentiable functions of x.
11
Step 1: Let’s choose u= ln(x)and dv =x2dx. Then, du =1
xdx and
v=1
3x3.
Step 2: Now, let’s apply the integration by parts formula:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C,
where Cis the constant of integration.
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C.
Question 15
Question
Evaluate the definite integral ∫1
0xsin−1(x)dx using integration by parts.
Solution
To evaluate the given definite integral ∫1
0xsin−1(x)dx using integration by
parts, we will use the formula ∫u dv =uv −∫v du.
Step 1: Choose uand dv
Let u= sin−1(x)and dv =x dx. Then, du =1
√1−x2dx and v=1
2x2.
Step 2: Apply integration by parts
∫xsin−1(x)dx =uv −∫v du
= sin−1(x)(1
2x2)−∫(1
2x2·1
√1−x2)dx
=1
2x2sin−1(x)−1
2∫x2
√1−x2dx
Step 3: Evaluate the remaining integral
To evaluate ∫x2
√1−x2dx, we can use the substitution u= 1−x2,du =−2x dx:
∫x2
√1−x2dx =−1
2∫−2x
√1−x2·x dx
=−1
2∫−du
√u
=√u+C
=√1−x2+C
12
Step 4: Substitute back and apply limits
∫1
0
xsin−1(x)dx =[1
2x2sin−1(x)−1
2(√1−x2)]1
0
=(1
2sin−1(1) −1
2√1−1)−(1
2sin−1(0) −1
2√1−0)
=π
4
Therefore, ∫1
0xsin−1(x)dx =π
4.
Question 16
Question
Evaluate the integral: ∫x2exdx.
Solution
To evaluate the given integral, we will use integration by parts. Integration by
parts is given by the formula:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let u=x2and dv =exdx. Then, calculate du and v:
du = 2x dx and v=∫exdx =ex
Step 2: Apply the formula for integration by parts:
∫x2exdx =x2ex−∫2xexdx
Step 3: Now, we need to evaluate the remaining integral on the right-hand
side, ∫2xexdx, using integration by parts again.
Step 4: Let u= 2xand dv =exdx. Calculate du and v:
du = 2 dx and v=ex
Step 5: Apply the formula for integration by parts to evaluate ∫2xexdx:
∫2xexdx = 2xex−∫2exdx
13
Step 6: Simplify the integral:
∫2exdx = 2 ∫exdx = 2ex
Step 7: Substitute back into our original integral:
∫x2exdx =x2ex−2xex+C
Therefore, ∫x2exdx =x2ex−2xex+C, where Cis the constant of integra-
tion.
Question 17
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx, we can use the integration by parts for-
mula: ∫u dv =uv −∫v du.
Let u= ln(x)and dv =x dx. Then, we have:
du =1
xdx
v=1
2x2
14
Now, we will apply the integration by parts formula:
Step 1: Let u= ln(x), dv =x dx
Step 2: Calculate du and v:
du =1
xdx, v =1
2x2
Step 3: Apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)(1
2x2)−∫(1
2x2)(1
x)dx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 18
Question
Compute the integral ∫x2exdx using integration by parts.
Solution
To evaluate the given integral ∫x2exdx using integration by parts, we will apply
the formula ∫u dv =uv −∫v du with u=x2and dv =exdx.
Step 1: Let’s determine du and v. We have:
du =d
dx (x2)dx = 2x dx
v=∫exdx =ex
Step 2: Apply integration by parts formula.
∫x2exdx =x2ex−∫ex·2x dx
=x2ex−2∫xexdx
15
Now, we need to apply integration by parts again to evaluate the remaining
integral ∫xexdx.
Step 3: Let’s apply integration by parts to ∫xexdx. Choose u=xand
dv =exdx.
Now we determine du and v:
du =dx
v=∫exdx =ex
Step 4: Apply integration by parts formula to ∫xexdx.
∫xexdx =xex−∫exdx
=xex−ex
Step 5: Substitute back into our original integral.
∫x2exdx =x2ex−2∫xexdx
=x2ex−2(xex−ex)
=x2ex−2xex+ 2ex+C
Therefore, ∫x2exdx =x2ex−2xex+ 2ex+C, where Cis the constant of
integration.
Question 19
Question
Evaluate the following integral:
∫x2cos x dx
Solution
To evaluate the given integral ∫x2cos x dx, we will use integration by parts.
Step 1: Let’s choose our uand dv. Let u=x2and dv = cos x dx.
Step 2: Compute du and v.
•du =d
dx (x2)dx = 2x dx
•v=∫cos x dx = sin x
16
Step 3: Apply the formula for integration by parts.
∫u dv =uv −∫v du
Now, we substitute u,dv,du, and vinto the formula:
∫x2cos x dx =x2sin x−∫sin x·2x dx
Step 4: Simplify the resulting integral.
∫x2cos x dx =x2sin x−2∫xsin x dx
Step 5: Let’s use integration by parts again for the remaining integral. Let
u=xand dv = sin x dx.
Now, compute du and v:
•du =d
dx (x)dx =dx
•v=∫sin x dx =−cos x
Step 6: Apply the formula for integration by parts again.
∫x2cos x dx =x2sin x−2(x(−cos x)−∫(−cos x)dx)
Simplify further to get the final result:
∫x2cos x dx =x2sin x+ 2xcos x−2 sin x+C
Where Cis the constant of integration.
Question 20
Question
Evaluate the integral ∫x2sin−1x dx using integration by parts.
Solution
To evaluate the given integral ∫x2sin−1x dx using integration by parts, we need
to choose two functions to assign as uand dv in the formula:
∫u dv =uv −∫v du
Let’s choose u= sin−1xand dv =x2dx. Then, we can find du and v:
du =1
√1−x2dx
17
v=1
3x3
Now, we can apply the integration by parts formula:
∫x2sin−1x dx =1
3x3sin−1x−∫1
3x3·1
√1−x2dx
∫x2sin−1x dx =1
3x3sin−1x−1
3∫x3
√1−x2dx
Integrating the remaining integral ∫x3
√1−x2dx would require further manip-
ulation using substitution or other techniques.
Question 21
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x2ln(x)dx, we will use integration by parts, which
states ∫u dv =uv −∫v du.
Let u= ln(x)and dv =x2dx. Then, we have
du =1
xdx and v=1
3x3.
Step 1: Apply integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx.
Step 2: Integrate the remaining integral:
∫x2dx =1
3x3+C,
where Cis the constant of integration.
Step 3: Substitute back to find the final answer:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
=1
3x3(ln(x)−1
3) + C.
Therefore, ∫x2ln(x)dx =1
3x3(ln(x)−1
3) + C.
18
Question 22
Question
Evaluate the integral
∫xe2xdx
Solution
To evaluate the given integral, we will use the technique of integration by parts,
which states ∫u dv =uv −∫v du
Let u=xand dv =e2xdx. Then, we have du =dx and v=1
2e2x.
Step 1: Apply integration by parts:
∫xe2xdx =x(1
2e2x)−∫(1
2e2x)dx
=1
2xe2x−1
4e2x+C
Step 2: Simplify the result:
∫xe2xdx =1
2xe2x−1
4e2x+C
Therefore, the integral ∫xe2xdx evaluates to 1
2xe2x−1
4e2x+C.
Question 23
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will have to
choose which part of the integrand to differentiate and which part to integrate.
Let’s follow the formula for integration by parts:
∫u dv =uv −∫v du
Step 1: Choose u= ln(x)and dv =x dx.
Then, differentiate uto get:
du =1
xdx
19
And integrate dv to get:
v=x2
2
Step 2: Apply the integration by parts formula:
∫xln(x)dx = ln(x)·x2
2−∫x2
2·1
xdx
Simplify the expression:
∫xln(x)dx =x2
2ln(x)−1
2∫x dx
Step 3: Integrate the remaining integral:
∫x dx =x2
2
Step 4: Substitute back into the equation:
∫xln(x)dx =x2
2ln(x)−1
2·x2
2+C
∫xln(x)dx =x2
2ln(x)−x2
4+C
Therefore, the result of the given integral is x2
2ln(x)−x2
4+C, where Cis
the constant of integration.
Question 24
Question
Evaluate the integral ∫excos(x)dx using integration by parts.
Solution
To evaluate the given integral ∫excos(x)dx using integration by parts, we will
apply the formula:
∫u dv =uv −∫v du
Step 1: Let’s choose uand dv:
u=exand dv = cos(x)dx
Step 2: Compute du and v:
du =exdx
20
v=∫cos(x)dx = sin(x)
Step 3: Apply the integration by parts formula:
∫excos(x)dx =exsin(x)−∫sin(x)·exdx
Step 4: Now we have a new integral to evaluate. Let’s again apply integra-
tion by parts with u= sin(x)and dv =exdx:
u= sin(x)and dv =exdx
du = cos(x)dx
v=∫exdx =ex
Step 5: Apply the integration by parts formula:
∫sin(x)·exdx =exsin(x)−∫excos(x)dx
Step 6: Substitute this back into the first equation:
∫excos(x)dx =exsin(x)−(exsin(x)−∫excos(x)dx)
Step 7: Simplify and solve for the original integral:
2∫excos(x)dx =exsin(x)
∫excos(x)dx =exsin(x)
2+C
Therefore, ∫excos(x)dx =exsin(x)
2+C, where Cis the constant of integra-
tion.
Question 25
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
21
Solution
To integrate ∫xln(x)dx, we will use integration by parts where we will set
u= ln(x)and dv =x dx.
Step 1: Compute du and v.
•u= ln(x)
•du =1
xdx
•dv =x dx =⇒v=1
2x2
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C(where Cis the constant of integration)
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C.
22
