MATH 132 - CALCULUS AND
ANALYTIC GEOMETRY II -
Integration by Parts
Question Bank - Set 5
Liberty University
Question 1
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 choose
u= ln(x)and dv =x dx.
Step 1: Compute du and v.
• Let u= ln(x)
• Compute du =1
xdx
• Integrate dv =x dx to get 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
Step 3: Evaluate the integral.
=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 2
Question
Evaluate the integral ∫exsin(x)dx using integration by parts.
Solution
To evaluate the integral ∫exsin(x)dx using integration by parts, we will choose
u=exand dv = sin(x)dx. Then, we’ll differentiate uto find du and integrate
dv to find v.
Step 1: Let u=exand dv = sin(x)dx. Then,
du =exdx and v=−cos(x).
Step 2: Now, we can use the integration by parts formula:
∫u dv =uv −∫v du.
Substitute u,v,du, and dv:
∫exsin(x)dx =ex(−cos(x)) −∫(−cos(x))(exdx).
Step 3: Simplify the equation:
=−excos(x) + ∫excos(x)dx.
Step 4: We need to evaluate the new integral ∫excos(x)dx using integra-
tion by parts again. Let u=exand dv = cos(x)dx. Then,
du =exdx and v= sin(x).
Step 5: Apply integration by parts with u=exand dv = cos(x)dx:
∫excos(x)dx =exsin(x)−∫sin(x)exdx.
Step 6: Substitute back the integral into our expression:
=−excos(x)+(exsin(x)−∫exsin(x)dx).
Step 7: Rearrange terms and isolate the original integral on one side:
2∫exsin(x)dx =ex(sin(x)−cos(x)).
Finally, solve for the integral:
∫exsin(x)dx =ex(sin(x)−cos(x))
2+C,
where Cis the constant of integration.
2
Question 3
Question
Evaluate the integral ∫xexsin x dx using integration by parts.
Solution
To evaluate the integral ∫xexsin x dx using integration by parts, we will apply
the formula ∫u dv =uv −∫v du with the appropriate choices for uand dv.
Step 1: Let us choose u=xand dv =exsin x dx. Then, we have:
du =dx and v=∫exsin x dx
Step 2: To find v, we can use integration by parts again by choosing u=
sin xand dv =exdx:
du = cos x dx and v=−excos x+∫excos x dx
Step 3: Now, we need to find ∫excos x dx. Let w=exand dz = cos x dx.
Then, we have:
dw =exdx and z=∫cos x dx = sin x
Step 4: Substituting wand zback, we get:
∫excos x dx =exsin x−∫exsin x dx
Step 5: Now, substituting vand the expression for ∫excos x dx back into
the integration by parts formula, we have:
∫xexsin x dx =uv −∫v du
=x(−excos x)−(−excos x+exsin x−∫exsin x dx)
=−xexcos x+excos x−exsin x+∫exsin x dx
Step 6: Rearranging terms, we get:
2∫xexsin x dx =ex(sin x−xcos x)
∫xexsin x dx =ex(sin x−xcos x)
2+C
So, the solution to the integral ∫xexsin x dx is ex(sin x−xcos x)
2+C, where C
is the constant of integration.
3
Question 4
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 choose
u=xand dv = cos(x)dx. Then, we will find du and vto apply the integration
by parts formula: ∫u dv =uv −∫v du.
Step 1: Find du and v.
u=x, dv = cos(x)dx
du =dx, v =∫cos(x)dx = sin(x)
Step 2: Apply the integration by parts formula.
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C
Therefore, ∫xcos(x)dx =xsin(x) + cos(x) + C, where Cis the constant of
integration.
Question 5
Question
Evaluate ∫xsin(x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(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: Choose uand dv. Let u=xand dv = sin(x)dx. Then, we have:
du =dx and v=−cos(x)
4
Step 2: Apply the integration by parts formula.
∫xsin(x)dx =uv −∫v du
=x(−cos(x)) −∫(−cos(x)dx)
=−xcos(x) + ∫cos(x)dx
Step 3: Integrate ∫cos(x)dx.
∫cos(x)dx = sin(x) + C
Step 4: Substitute back into the equation.
∫xsin(x)dx =−xcos(x) + sin(x) + C
Therefore, ∫xsin(x)dx =−xcos(x) + sin(x) + C, where Cis the constant
of integration.
Question 6
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 follow
the formula:
∫u dv =uv −∫v du
where we choose uand dv appropriately. Let us take u= ln(x)and dv =
x dx.
Step 1: Calculate du and v.
•du =1
xdx
• To find v, we integrate dv:
v=∫x dx =x2
2
5
Step 2: Apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)·x2
2−∫x2
2·1
xdx
=x2ln(x)
2−1
2∫x dx
=x2ln(x)
2−x2
4+C
Therefore, ∫xln(x)dx =x2ln(x)
2−x2
4+C, where Cis the constant of
integration.
Question 7
Question
Evaluate the integral ∫x3ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x3ln(x)dx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let’s choose:
u= ln(x)⇒du =1
xdx
dv =x3dx ⇒v=1
4x4
Step 2: Now, we can apply the integration by parts formula:
∫x3ln(x)dx = ln(x)·1
4x4−∫1
4x4·1
xdx
Step 3: Simplify the integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4∫x3dx
Step 4: Evaluate the remaining integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4·1
4x4+C
6
∫x3ln(x)dx =1
4x4ln(x)−1
16x4+C
Therefore, the integral ∫x3ln(x)dx simplifies to 1
4x4ln(x)−1
16 x4+C, where
Cis the constant of integration.
Question 8
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 choose
u= ln(x)and dv =x2dx.
Step 1: Calculate du and v:
u= ln(x)
dv =x2dx
du =1
xdx
v=1
3x3
Step 2: Apply the formula for integration by parts:
∫x2ln(x)dx =uv −∫v du
=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.
7
Solution
To evaluate ∫xln(x)dx using integration by parts, we will use the formula:
∫u dv =uv −∫v du
Let’s choose u= ln(x)and dv =x dx. Then, we have:
du =1
xdx and v=1
2x2
Step 1: Calculate du and v.
du =1
xdx ⇒du =dx
v=1
2x2
Step 2: Apply integration by parts formula.
∫xln(x)dx =∫u dv
=uv −∫v du
= ln(x)·1
2x2−∫1
2x2dx
=1
2x2ln(x)−1
2∫x2dx
Step 3: Evaluate ∫x2dx.
∫x2dx =1
3x3+C
Step 4: Substitute back into the original integral.
∫xln(x)dx =1
2x2ln(x)−1
2·1
3x3+C
=1
2x2ln(x)−1
6x3+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
6x3+C, where Cis the constant of
integration.
Question 10
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
8
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du.
Step 1: Choose uand dv
Let u= ln(x)and dv =x dx. Then, we have: du =1
xdx and v=1
2x2.
Step 2: Compute du and v
Now, we have: du =1
xdx and v=1
2x2.
Step 3: Apply integration by parts formula
Using the formula ∫u dv =uv −∫v du, we get:
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
Step 4: Simplify the expression
Simplifying, we get: 1
2x2ln(x)−1
2∫x dx
Step 5: Integrate the new integral
Integrating ∫x dx =1
2x2, we have:
1
2x2ln(x)−1
4x2+C
Step 6: Final answer
Therefore, the integral ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the
constant of integration.
Question 11
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 use the
formula ∫u dv =uv −∫v du where uand vare differentiable functions of x.
Step 1: Let’s choose u= ln xand dv =x2dx. Then we have:
du =1
xdx and v=1
3x3
9
Step 2: Now, we apply the integration by parts formula:
∫x2ln x dx =uv −∫v du
= ln x·1
3x3−∫1
3x3·1
xdx
=1
3x3ln x−1
3∫x2dx
=1
3x3ln x−1
3·1
3x3+C
=1
3x3ln x−1
9x3+C
Step 3: Therefore, the integral ∫x2ln x dx evaluates to 1
3x3ln x−1
9x3+C,
where Cis the constant of integration.
Question 12
Question
Evaluate the integral ∫x2ln x dx using integration by parts.
Solution
To evaluate the integral ∫x2ln x dx using integration by parts, we need to
choose which part of the integrand to differentiate (letting this be u) and which
part to integrate (letting this be dv).
Let u= ln xand dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Now, we use the formula for integration by parts:
∫u dv =uv −∫v du
Step 1: Compute uv.
ln x·1
3x3=1
3x3ln x
Step 2: Compute ∫v du.
∫1
3x3·1
xdx =1
3∫x2dx =1
3·1
3x3+C=1
9x3+C
Therefore, the integral of ∫x2ln x dx is:
∫x2ln x dx =1
3x3ln x−1
9x3+C
where Cis the constant of integration.
10
Question 13
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we can choose
u= ln(x)and dv =x dx.
Step 1: Compute du and v.
• Let u= ln(x). Then, du =1
xdx.
• Let dv =x dx. Integrating dv gives v=1
2x2.
Step 2: Apply the integration by parts formula ∫u dv =uv −∫v du.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 14
Question
Evaluate the integral ∫xexdx.
Solution
To evaluate the integral ∫xexdx, we will use integration by parts. Integration
by parts is based on the formula ∫u dv =uv −∫v du, where uand vare
differentiable functions of x.
Step 1: Choose uand dv. Let u=xand dv =exdx.
Step 2: Compute du and v. We have du =dx and v=∫exdx =ex.
Step 3: Apply the integration by parts formula.
∫xexdx =uv −∫v du
∫xexdx =x·ex−∫exdx
11
Step 4: Simplify the expression.
∫xexdx =x·ex−ex+C
Therefore, ∫xexdx =xex−ex+C, where Cis the constant of integration.
Question 15
Question
Evaluate the integral ∫x3ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x3ln(x)dx using integration by parts, we will use the
formula ∫u dv =uv −∫v du. Let u= ln(x)and dv =x3dx, then du =1
xdx
and v=1
4x4.
Step 1: Apply integration by parts:
∫x3ln(x)dx =1
4x4ln(x)−∫1
4x4·1
xdx
Step 2: Simplify the integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4∫x3dx
Step 3: Evaluate the remaining integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4·1
4x4+C
Step 4: Simplify the expression:
∫x3ln(x)dx =1
4x4ln(x)−1
16x4+C
Therefore, ∫x3ln(x)dx =1
4x4ln(x)−1
16 x4+C, where Cis the constant of
integration.
Question 16
Question
Evaluate the integral
∫xcos(2x)dx
12
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 we choose uand dv appropriately.
Step 1: Let’s choose u=xand dv = cos(2x)dx. Then, we have:
du =dx and v=1
2sin(2x)
Step 2: Now, applying the integration by parts formula:
∫xcos(2x)dx =uv −∫v du
=x·1
2sin(2x)−∫1
2sin(2x)dx
=1
2xsin(2x)−1
2∫sin(2x)dx
Step 3: Continuing to solve the integral:
∫xcos(2x)dx =1
2xsin(2x)−1
2(−1
2cos(2x)) + C
=1
2xsin(2x) + 1
4cos(2x) + C
Therefore, the solution to the given integral is
∫xcos(2x)dx =1
2xsin(2x) + 1
4cos(2x) + C
where Cis the constant of integration.
Question 17
Question
Evaluate the integral ∫xln x dx using integration by parts.
Solution
To evaluate the integral ∫xln x dx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
13
where uand vare differentiable functions of x.
Step 1: Identify uand dv Let u= ln xand dv =x dx. Then,
du =1
xdx and v=∫x dx =1
2x2
Step 2: Apply integration by parts Using the integration by parts
formula, we have:
∫xln x dx =uv −∫v du = ln x·1
2x2−∫1
2x2·1
xdx
Step 3: Simplify the integral Simplifying the expression gives:
1
2x2ln x−1
2∫x dx
Step 4: Evaluate the integral Now, we can integrate ∫x dx to get:
1
2x2ln x−1
2·1
2x2+C
Step 5: Final answer Therefore, the solution to the integral ∫xln x dx is:
1
2x2ln x−1
4x2+C
where Cis the constant of integration.
Question 18
Question
Evaluate the integral: ∫x2ln(x)dx.
Solution
To evaluate the given integral, we can use integration by parts. Let’s choose
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
Step 2: Simplify the integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3∫x2dx
Step 3: Integrate the remaining term:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
Therefore, the solution to the given integral is 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
14
Question 19
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To solve the integral ∫xln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x dx.
Step 1: Calculate 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
Step 3: Integrate ∫x dx.
∫x dx =1
2x2+C
Step 4: Substitute back into the previous expression.
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 20
Question
Evaluate the integral ∫xarcsin(x)dx.
15
Solution
To evaluate the integral ∫xarcsin(x)dx, we will use integration by parts where
we let u= arcsin(x)and dv =x dx. Then, we will find du and v.
•Step 1: Find du by taking the derivative of u.
u= arcsin(x)
du
dx =1
√1−x2
du =1
√1−x2dx
•Step 2: Find vby integrating dv.
dv =x dx
v=x2
2
•Step 3: Apply integration by parts formula:
∫u dv =uv −∫v du
= arcsin(x)x2
2−∫x2
2·1
√1−x2dx
•Step 4: Integrate ∫x2
2·1
√1−x2dx using the substitution method x=
sin(θ).
∫x2
2·1
√1−x2dx =∫sin2(θ)
2·1
cos(θ)·cos(θ)dθ
=1
2∫sin2(θ)dθ
=1
2∫1−cos(2θ)
2dθ
=1
4(θ−sin(2θ)
2)+C
=1
4(arcsin(x)−x√1−x2)+C
•Step 5: Substitute back xusing the first substitution x= sin(θ)in the
original integral:
∫xarcsin(x)dx = arcsin(x)x2
2−1
4(arcsin(x)−x√1−x2)+C
=1
2x2arcsin(x)−1
4arcsin(x) + 1
4x√1−x2+C
16
Question 21
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral using integration by parts, we need to choose
which part of the integrand to differentiate and which part to integrate. Let’s
follow the integration by parts formula, which states:
∫u dv =uv −∫v du.
Step 1: Choose u= ln(x)and dv =x2dx. Then, calculate du and v:
•du =1
xdx
•v=∫x2dx =x3
3
Step 2: Apply the integration by parts formula:
∫x2ln(x)dx =x3ln(x)
3−∫x3
3·1
xdx.
Step 3: Simplify the integral:
∫x2ln(x)dx =x3ln(x)
3−∫x2
3dx
=x3ln(x)
3−1
3∫x2dx
=x3ln(x)
3−x3
9+C,
where Cis the constant of integration.
Therefore, the integral of x2ln(x)is x3ln(x)
3−x3
9+C.
Question 22
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
17
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x2dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Let’s find du and v.
•u= ln(x)
•du =1
xdx
•dv =x2dx =⇒v=x3
3
Step 2: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫x2ln(x)dx =x3
3ln(x)−∫x3
3·1
xdx
=x3
3ln(x)−1
3∫x2dx
=x3
3ln(x)−1
3·x3
3+C
=x3
3ln(x)−x3
9+C
Therefore, ∫x2ln(x)dx =x3
3ln(x)−x3
9+C, where Cis the constant of
integration.
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 choose:
u= ln(x)
dv =x dx
Step 1: Compute du and v.
du =1
xdx
v=1
2x2
18
Step 2: Apply the integration by parts formula:
∫u dv =uv −∫v du
∫xln(x)dx =1
2x2ln(x)−∫1
2x dx
Step 3: Integrate ∫1
2x dx.
1
2∫x dx =1
2·1
2x2=1
4x2
Step 4: Substitute 1
4x2back into the equation:
1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 24
Question
Evaluate the definite integral:
∫1
0
xsin−1(x)dx
Solution
Step 1: Let u= sin−1(x)and dv =x dx. Then, du =1
√1−x2dx and v=x2
2.
Step 2: Now, we can apply the integration by parts formula:
∫u dv =uv −∫v du
Step 3: Substituting the values of u,v,du, and dv into the formula, we get:
∫1
0
xsin−1(x)dx =[x2
2sin−1(x)]1
0−∫1
0
x2
2·1
√1−x2dx
Step 4: We need to evaluate the integral on the right-hand side. Let’s make
a substitution x= sin(θ), then dx = cos(θ)dθ.
Step 5: After the substitution, the integral becomes:
∫π
2
0
sin2(θ)
2dθ =1
2∫π
2
0
1−cos(2θ)
2dθ
19
Step 6: Now, we can split the integral into two parts and evaluate them
separately:
1
4[θ−sin(2θ)
2]
π
2
0
=π
4
Step 7: Finally, substituting back the values of the definite integral and
simplifying, we get:
∫1
0
xsin−1(x)dx =1
2sin−1(1) −π
4=π
4−π
4= 0
Question 25
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 choose
u= ln(x)and dv =x dx.
Step 1: Calculate du and v.
•u= ln(x)
•du =1
xdx
•v=∫x dx =1
2x2
Step 2: Apply 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
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.
20
Question 2
Question
Evaluate the integral ∫exsin(x)dx using integration by parts.
Solution
To evaluate the integral ∫exsin(x)dx using integration by parts, we will choose
u=exand dv = sin(x)dx. Then, we’ll differentiate uto find du and integrate
dv to find v.
Step 1: Let u=exand dv = sin(x)dx. Then,
du =exdx and v=−cos(x).
Step 2: Now, we can use the integration by parts formula:
∫u dv =uv −∫v du.
Substitute u,v,du, and dv:
∫exsin(x)dx =ex(−cos(x)) −∫(−cos(x))(exdx).
Step 3: Simplify the equation:
=−excos(x) + ∫excos(x)dx.
Step 4: We need to evaluate the new integral ∫excos(x)dx using integra-
tion by parts again. Let u=exand dv = cos(x)dx. Then,
du =exdx and v= sin(x).
Step 5: Apply integration by parts with u=exand dv = cos(x)dx:
∫excos(x)dx =exsin(x)−∫sin(x)exdx.
Step 6: Substitute back the integral into our expression:
=−excos(x)+(exsin(x)−∫exsin(x)dx).
Step 7: Rearrange terms and isolate the original integral on one side:
2∫exsin(x)dx =ex(sin(x)−cos(x)).
Finally, solve for the integral:
∫exsin(x)dx =ex(sin(x)−cos(x))
2+C,
where Cis the constant of integration.
2
Question 3
Question
Evaluate the integral ∫xexsin x dx using integration by parts.
Solution
To evaluate the integral ∫xexsin x dx using integration by parts, we will apply
the formula ∫u dv =uv −∫v du with the appropriate choices for uand dv.
Step 1: Let us choose u=xand dv =exsin x dx. Then, we have:
du =dx and v=∫exsin x dx
Step 2: To find v, we can use integration by parts again by choosing u=
sin xand dv =exdx:
du = cos x dx and v=−excos x+∫excos x dx
Step 3: Now, we need to find ∫excos x dx. Let w=exand dz = cos x dx.
Then, we have:
dw =exdx and z=∫cos x dx = sin x
Step 4: Substituting wand zback, we get:
∫excos x dx =exsin x−∫exsin x dx
Step 5: Now, substituting vand the expression for ∫excos x dx back into
the integration by parts formula, we have:
∫xexsin x dx =uv −∫v du
=x(−excos x)−(−excos x+exsin x−∫exsin x dx)
=−xexcos x+excos x−exsin x+∫exsin x dx
Step 6: Rearranging terms, we get:
2∫xexsin x dx =ex(sin x−xcos x)
∫xexsin x dx =ex(sin x−xcos x)
2+C
So, the solution to the integral ∫xexsin x dx is ex(sin x−xcos x)
2+C, where C
is the constant of integration.
3
Question 4
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 choose
u=xand dv = cos(x)dx. Then, we will find du and vto apply the integration
by parts formula: ∫u dv =uv −∫v du.
Step 1: Find du and v.
u=x, dv = cos(x)dx
du =dx, v =∫cos(x)dx = sin(x)
Step 2: Apply the integration by parts formula.
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C
Therefore, ∫xcos(x)dx =xsin(x) + cos(x) + C, where Cis the constant of
integration.
Question 5
Question
Evaluate ∫xsin(x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(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: Choose uand dv. Let u=xand dv = sin(x)dx. Then, we have:
du =dx and v=−cos(x)
4
Step 2: Apply the integration by parts formula.
∫xsin(x)dx =uv −∫v du
=x(−cos(x)) −∫(−cos(x)dx)
=−xcos(x) + ∫cos(x)dx
Step 3: Integrate ∫cos(x)dx.
∫cos(x)dx = sin(x) + C
Step 4: Substitute back into the equation.
∫xsin(x)dx =−xcos(x) + sin(x) + C
Therefore, ∫xsin(x)dx =−xcos(x) + sin(x) + C, where Cis the constant
of integration.
Question 6
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 follow
the formula:
∫u dv =uv −∫v du
where we choose uand dv appropriately. Let us take u= ln(x)and dv =
x dx.
Step 1: Calculate du and v.
•du =1
xdx
• To find v, we integrate dv:
v=∫x dx =x2
2
5
Step 2: Apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)·x2
2−∫x2
2·1
xdx
=x2ln(x)
2−1
2∫x dx
=x2ln(x)
2−x2
4+C
Therefore, ∫xln(x)dx =x2ln(x)
2−x2
4+C, where Cis the constant of
integration.
Question 7
Question
Evaluate the integral ∫x3ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x3ln(x)dx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let’s choose:
u= ln(x)⇒du =1
xdx
dv =x3dx ⇒v=1
4x4
Step 2: Now, we can apply the integration by parts formula:
∫x3ln(x)dx = ln(x)·1
4x4−∫1
4x4·1
xdx
Step 3: Simplify the integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4∫x3dx
Step 4: Evaluate the remaining integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4·1
4x4+C
6
∫x3ln(x)dx =1
4x4ln(x)−1
16x4+C
Therefore, the integral ∫x3ln(x)dx simplifies to 1
4x4ln(x)−1
16 x4+C, where
Cis the constant of integration.
Question 8
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 choose
u= ln(x)and dv =x2dx.
Step 1: Calculate du and v:
u= ln(x)
dv =x2dx
du =1
xdx
v=1
3x3
Step 2: Apply the formula for integration by parts:
∫x2ln(x)dx =uv −∫v du
=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.
7
Solution
To evaluate ∫xln(x)dx using integration by parts, we will use the formula:
∫u dv =uv −∫v du
Let’s choose u= ln(x)and dv =x dx. Then, we have:
du =1
xdx and v=1
2x2
Step 1: Calculate du and v.
du =1
xdx ⇒du =dx
v=1
2x2
Step 2: Apply integration by parts formula.
∫xln(x)dx =∫u dv
=uv −∫v du
= ln(x)·1
2x2−∫1
2x2dx
=1
2x2ln(x)−1
2∫x2dx
Step 3: Evaluate ∫x2dx.
∫x2dx =1
3x3+C
Step 4: Substitute back into the original integral.
∫xln(x)dx =1
2x2ln(x)−1
2·1
3x3+C
=1
2x2ln(x)−1
6x3+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
6x3+C, where Cis the constant of
integration.
Question 10
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
8
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du.
Step 1: Choose uand dv
Let u= ln(x)and dv =x dx. Then, we have: du =1
xdx and v=1
2x2.
Step 2: Compute du and v
Now, we have: du =1
xdx and v=1
2x2.
Step 3: Apply integration by parts formula
Using the formula ∫u dv =uv −∫v du, we get:
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
Step 4: Simplify the expression
Simplifying, we get: 1
2x2ln(x)−1
2∫x dx
Step 5: Integrate the new integral
Integrating ∫x dx =1
2x2, we have:
1
2x2ln(x)−1
4x2+C
Step 6: Final answer
Therefore, the integral ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the
constant of integration.
Question 11
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 use the
formula ∫u dv =uv −∫v du where uand vare differentiable functions of x.
Step 1: Let’s choose u= ln xand dv =x2dx. Then we have:
du =1
xdx and v=1
3x3
9
Step 2: Now, we apply the integration by parts formula:
∫x2ln x dx =uv −∫v du
= ln x·1
3x3−∫1
3x3·1
xdx
=1
3x3ln x−1
3∫x2dx
=1
3x3ln x−1
3·1
3x3+C
=1
3x3ln x−1
9x3+C
Step 3: Therefore, the integral ∫x2ln x dx evaluates to 1
3x3ln x−1
9x3+C,
where Cis the constant of integration.
Question 12
Question
Evaluate the integral ∫x2ln x dx using integration by parts.
Solution
To evaluate the integral ∫x2ln x dx using integration by parts, we need to
choose which part of the integrand to differentiate (letting this be u) and which
part to integrate (letting this be dv).
Let u= ln xand dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Now, we use the formula for integration by parts:
∫u dv =uv −∫v du
Step 1: Compute uv.
ln x·1
3x3=1
3x3ln x
Step 2: Compute ∫v du.
∫1
3x3·1
xdx =1
3∫x2dx =1
3·1
3x3+C=1
9x3+C
Therefore, the integral of ∫x2ln x dx is:
∫x2ln x dx =1
3x3ln x−1
9x3+C
where Cis the constant of integration.
10
Question 13
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we can choose
u= ln(x)and dv =x dx.
Step 1: Compute du and v.
• Let u= ln(x). Then, du =1
xdx.
• Let dv =x dx. Integrating dv gives v=1
2x2.
Step 2: Apply the integration by parts formula ∫u dv =uv −∫v du.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 14
Question
Evaluate the integral ∫xexdx.
Solution
To evaluate the integral ∫xexdx, we will use integration by parts. Integration
by parts is based on the formula ∫u dv =uv −∫v du, where uand vare
differentiable functions of x.
Step 1: Choose uand dv. Let u=xand dv =exdx.
Step 2: Compute du and v. We have du =dx and v=∫exdx =ex.
Step 3: Apply the integration by parts formula.
∫xexdx =uv −∫v du
∫xexdx =x·ex−∫exdx
11
Step 4: Simplify the expression.
∫xexdx =x·ex−ex+C
Therefore, ∫xexdx =xex−ex+C, where Cis the constant of integration.
Question 15
Question
Evaluate the integral ∫x3ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x3ln(x)dx using integration by parts, we will use the
formula ∫u dv =uv −∫v du. Let u= ln(x)and dv =x3dx, then du =1
xdx
and v=1
4x4.
Step 1: Apply integration by parts:
∫x3ln(x)dx =1
4x4ln(x)−∫1
4x4·1
xdx
Step 2: Simplify the integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4∫x3dx
Step 3: Evaluate the remaining integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4·1
4x4+C
Step 4: Simplify the expression:
∫x3ln(x)dx =1
4x4ln(x)−1
16x4+C
Therefore, ∫x3ln(x)dx =1
4x4ln(x)−1
16 x4+C, where Cis the constant of
integration.
Question 16
Question
Evaluate the integral
∫xcos(2x)dx
12
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 we choose uand dv appropriately.
Step 1: Let’s choose u=xand dv = cos(2x)dx. Then, we have:
du =dx and v=1
2sin(2x)
Step 2: Now, applying the integration by parts formula:
∫xcos(2x)dx =uv −∫v du
=x·1
2sin(2x)−∫1
2sin(2x)dx
=1
2xsin(2x)−1
2∫sin(2x)dx
Step 3: Continuing to solve the integral:
∫xcos(2x)dx =1
2xsin(2x)−1
2(−1
2cos(2x)) + C
=1
2xsin(2x) + 1
4cos(2x) + C
Therefore, the solution to the given integral is
∫xcos(2x)dx =1
2xsin(2x) + 1
4cos(2x) + C
where Cis the constant of integration.
Question 17
Question
Evaluate the integral ∫xln x dx using integration by parts.
Solution
To evaluate the integral ∫xln x dx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
13
where uand vare differentiable functions of x.
Step 1: Identify uand dv Let u= ln xand dv =x dx. Then,
du =1
xdx and v=∫x dx =1
2x2
Step 2: Apply integration by parts Using the integration by parts
formula, we have:
∫xln x dx =uv −∫v du = ln x·1
2x2−∫1
2x2·1
xdx
Step 3: Simplify the integral Simplifying the expression gives:
1
2x2ln x−1
2∫x dx
Step 4: Evaluate the integral Now, we can integrate ∫x dx to get:
1
2x2ln x−1
2·1
2x2+C
Step 5: Final answer Therefore, the solution to the integral ∫xln x dx is:
1
2x2ln x−1
4x2+C
where Cis the constant of integration.
Question 18
Question
Evaluate the integral: ∫x2ln(x)dx.
Solution
To evaluate the given integral, we can use integration by parts. Let’s choose
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
Step 2: Simplify the integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3∫x2dx
Step 3: Integrate the remaining term:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
Therefore, the solution to the given integral is 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
14
Question 19
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To solve the integral ∫xln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x dx.
Step 1: Calculate 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
Step 3: Integrate ∫x dx.
∫x dx =1
2x2+C
Step 4: Substitute back into the previous expression.
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 20
Question
Evaluate the integral ∫xarcsin(x)dx.
15
Solution
To evaluate the integral ∫xarcsin(x)dx, we will use integration by parts where
we let u= arcsin(x)and dv =x dx. Then, we will find du and v.
•Step 1: Find du by taking the derivative of u.
u= arcsin(x)
du
dx =1
√1−x2
du =1
√1−x2dx
•Step 2: Find vby integrating dv.
dv =x dx
v=x2
2
•Step 3: Apply integration by parts formula:
∫u dv =uv −∫v du
= arcsin(x)x2
2−∫x2
2·1
√1−x2dx
•Step 4: Integrate ∫x2
2·1
√1−x2dx using the substitution method x=
sin(θ).
∫x2
2·1
√1−x2dx =∫sin2(θ)
2·1
cos(θ)·cos(θ)dθ
=1
2∫sin2(θ)dθ
=1
2∫1−cos(2θ)
2dθ
=1
4(θ−sin(2θ)
2)+C
=1
4(arcsin(x)−x√1−x2)+C
•Step 5: Substitute back xusing the first substitution x= sin(θ)in the
original integral:
∫xarcsin(x)dx = arcsin(x)x2
2−1
4(arcsin(x)−x√1−x2)+C
=1
2x2arcsin(x)−1
4arcsin(x) + 1
4x√1−x2+C
16
Question 21
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral using integration by parts, we need to choose
which part of the integrand to differentiate and which part to integrate. Let’s
follow the integration by parts formula, which states:
∫u dv =uv −∫v du.
Step 1: Choose u= ln(x)and dv =x2dx. Then, calculate du and v:
•du =1
xdx
•v=∫x2dx =x3
3
Step 2: Apply the integration by parts formula:
∫x2ln(x)dx =x3ln(x)
3−∫x3
3·1
xdx.
Step 3: Simplify the integral:
∫x2ln(x)dx =x3ln(x)
3−∫x2
3dx
=x3ln(x)
3−1
3∫x2dx
=x3ln(x)
3−x3
9+C,
where Cis the constant of integration.
Therefore, the integral of x2ln(x)is x3ln(x)
3−x3
9+C.
Question 22
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
17
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x2dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Let’s find du and v.
•u= ln(x)
•du =1
xdx
•dv =x2dx =⇒v=x3
3
Step 2: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫x2ln(x)dx =x3
3ln(x)−∫x3
3·1
xdx
=x3
3ln(x)−1
3∫x2dx
=x3
3ln(x)−1
3·x3
3+C
=x3
3ln(x)−x3
9+C
Therefore, ∫x2ln(x)dx =x3
3ln(x)−x3
9+C, where Cis the constant of
integration.
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 choose:
u= ln(x)
dv =x dx
Step 1: Compute du and v.
du =1
xdx
v=1
2x2
18
Step 2: Apply the integration by parts formula:
∫u dv =uv −∫v du
∫xln(x)dx =1
2x2ln(x)−∫1
2x dx
Step 3: Integrate ∫1
2x dx.
1
2∫x dx =1
2·1
2x2=1
4x2
Step 4: Substitute 1
4x2back into the equation:
1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 24
Question
Evaluate the definite integral:
∫1
0
xsin−1(x)dx
Solution
Step 1: Let u= sin−1(x)and dv =x dx. Then, du =1
√1−x2dx and v=x2
2.
Step 2: Now, we can apply the integration by parts formula:
∫u dv =uv −∫v du
Step 3: Substituting the values of u,v,du, and dv into the formula, we get:
∫1
0
xsin−1(x)dx =[x2
2sin−1(x)]1
0−∫1
0
x2
2·1
√1−x2dx
Step 4: We need to evaluate the integral on the right-hand side. Let’s make
a substitution x= sin(θ), then dx = cos(θ)dθ.
Step 5: After the substitution, the integral becomes:
∫π
2
0
sin2(θ)
2dθ =1
2∫π
2
0
1−cos(2θ)
2dθ
19
Step 6: Now, we can split the integral into two parts and evaluate them
separately:
1
4[θ−sin(2θ)
2]
π
2
0
=π
4
Step 7: Finally, substituting back the values of the definite integral and
simplifying, we get:
∫1
0
xsin−1(x)dx =1
2sin−1(1) −π
4=π
4−π
4= 0
Question 25
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 choose
u= ln(x)and dv =x dx.
Step 1: Calculate du and v.
•u= ln(x)
•du =1
xdx
•v=∫x dx =1
2x2
Step 2: Apply 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
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.
20
Question 2
Question
Evaluate the integral ∫exsin(x)dx using integration by parts.
Solution
To evaluate the integral ∫exsin(x)dx using integration by parts, we will choose
u=exand dv = sin(x)dx. Then, we’ll differentiate uto find du and integrate
dv to find v.
Step 1: Let u=exand dv = sin(x)dx. Then,
du =exdx and v=−cos(x).
Step 2: Now, we can use the integration by parts formula:
∫u dv =uv −∫v du.
Substitute u,v,du, and dv:
∫exsin(x)dx =ex(−cos(x)) −∫(−cos(x))(exdx).
Step 3: Simplify the equation:
=−excos(x) + ∫excos(x)dx.
Step 4: We need to evaluate the new integral ∫excos(x)dx using integra-
tion by parts again. Let u=exand dv = cos(x)dx. Then,
du =exdx and v= sin(x).
Step 5: Apply integration by parts with u=exand dv = cos(x)dx:
∫excos(x)dx =exsin(x)−∫sin(x)exdx.
Step 6: Substitute back the integral into our expression:
=−excos(x)+(exsin(x)−∫exsin(x)dx).
Step 7: Rearrange terms and isolate the original integral on one side:
2∫exsin(x)dx =ex(sin(x)−cos(x)).
Finally, solve for the integral:
∫exsin(x)dx =ex(sin(x)−cos(x))
2+C,
where Cis the constant of integration.
2
Question 3
Question
Evaluate the integral ∫xexsin x dx using integration by parts.
Solution
To evaluate the integral ∫xexsin x dx using integration by parts, we will apply
the formula ∫u dv =uv −∫v du with the appropriate choices for uand dv.
Step 1: Let us choose u=xand dv =exsin x dx. Then, we have:
du =dx and v=∫exsin x dx
Step 2: To find v, we can use integration by parts again by choosing u=
sin xand dv =exdx:
du = cos x dx and v=−excos x+∫excos x dx
Step 3: Now, we need to find ∫excos x dx. Let w=exand dz = cos x dx.
Then, we have:
dw =exdx and z=∫cos x dx = sin x
Step 4: Substituting wand zback, we get:
∫excos x dx =exsin x−∫exsin x dx
Step 5: Now, substituting vand the expression for ∫excos x dx back into
the integration by parts formula, we have:
∫xexsin x dx =uv −∫v du
=x(−excos x)−(−excos x+exsin x−∫exsin x dx)
=−xexcos x+excos x−exsin x+∫exsin x dx
Step 6: Rearranging terms, we get:
2∫xexsin x dx =ex(sin x−xcos x)
∫xexsin x dx =ex(sin x−xcos x)
2+C
So, the solution to the integral ∫xexsin x dx is ex(sin x−xcos x)
2+C, where C
is the constant of integration.
3
Question 4
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 choose
u=xand dv = cos(x)dx. Then, we will find du and vto apply the integration
by parts formula: ∫u dv =uv −∫v du.
Step 1: Find du and v.
u=x, dv = cos(x)dx
du =dx, v =∫cos(x)dx = sin(x)
Step 2: Apply the integration by parts formula.
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C
Therefore, ∫xcos(x)dx =xsin(x) + cos(x) + C, where Cis the constant of
integration.
Question 5
Question
Evaluate ∫xsin(x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(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: Choose uand dv. Let u=xand dv = sin(x)dx. Then, we have:
du =dx and v=−cos(x)
4
Step 2: Apply the integration by parts formula.
∫xsin(x)dx =uv −∫v du
=x(−cos(x)) −∫(−cos(x)dx)
=−xcos(x) + ∫cos(x)dx
Step 3: Integrate ∫cos(x)dx.
∫cos(x)dx = sin(x) + C
Step 4: Substitute back into the equation.
∫xsin(x)dx =−xcos(x) + sin(x) + C
Therefore, ∫xsin(x)dx =−xcos(x) + sin(x) + C, where Cis the constant
of integration.
Question 6
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 follow
the formula:
∫u dv =uv −∫v du
where we choose uand dv appropriately. Let us take u= ln(x)and dv =
x dx.
Step 1: Calculate du and v.
•du =1
xdx
• To find v, we integrate dv:
v=∫x dx =x2
2
5
Step 2: Apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)·x2
2−∫x2
2·1
xdx
=x2ln(x)
2−1
2∫x dx
=x2ln(x)
2−x2
4+C
Therefore, ∫xln(x)dx =x2ln(x)
2−x2
4+C, where Cis the constant of
integration.
Question 7
Question
Evaluate the integral ∫x3ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x3ln(x)dx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let’s choose:
u= ln(x)⇒du =1
xdx
dv =x3dx ⇒v=1
4x4
Step 2: Now, we can apply the integration by parts formula:
∫x3ln(x)dx = ln(x)·1
4x4−∫1
4x4·1
xdx
Step 3: Simplify the integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4∫x3dx
Step 4: Evaluate the remaining integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4·1
4x4+C
6
∫x3ln(x)dx =1
4x4ln(x)−1
16x4+C
Therefore, the integral ∫x3ln(x)dx simplifies to 1
4x4ln(x)−1
16 x4+C, where
Cis the constant of integration.
Question 8
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 choose
u= ln(x)and dv =x2dx.
Step 1: Calculate du and v:
u= ln(x)
dv =x2dx
du =1
xdx
v=1
3x3
Step 2: Apply the formula for integration by parts:
∫x2ln(x)dx =uv −∫v du
=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.
7
Solution
To evaluate ∫xln(x)dx using integration by parts, we will use the formula:
∫u dv =uv −∫v du
Let’s choose u= ln(x)and dv =x dx. Then, we have:
du =1
xdx and v=1
2x2
Step 1: Calculate du and v.
du =1
xdx ⇒du =dx
v=1
2x2
Step 2: Apply integration by parts formula.
∫xln(x)dx =∫u dv
=uv −∫v du
= ln(x)·1
2x2−∫1
2x2dx
=1
2x2ln(x)−1
2∫x2dx
Step 3: Evaluate ∫x2dx.
∫x2dx =1
3x3+C
Step 4: Substitute back into the original integral.
∫xln(x)dx =1
2x2ln(x)−1
2·1
3x3+C
=1
2x2ln(x)−1
6x3+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
6x3+C, where Cis the constant of
integration.
Question 10
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
8
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du.
Step 1: Choose uand dv
Let u= ln(x)and dv =x dx. Then, we have: du =1
xdx and v=1
2x2.
Step 2: Compute du and v
Now, we have: du =1
xdx and v=1
2x2.
Step 3: Apply integration by parts formula
Using the formula ∫u dv =uv −∫v du, we get:
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
Step 4: Simplify the expression
Simplifying, we get: 1
2x2ln(x)−1
2∫x dx
Step 5: Integrate the new integral
Integrating ∫x dx =1
2x2, we have:
1
2x2ln(x)−1
4x2+C
Step 6: Final answer
Therefore, the integral ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the
constant of integration.
Question 11
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 use the
formula ∫u dv =uv −∫v du where uand vare differentiable functions of x.
Step 1: Let’s choose u= ln xand dv =x2dx. Then we have:
du =1
xdx and v=1
3x3
9
Step 2: Now, we apply the integration by parts formula:
∫x2ln x dx =uv −∫v du
= ln x·1
3x3−∫1
3x3·1
xdx
=1
3x3ln x−1
3∫x2dx
=1
3x3ln x−1
3·1
3x3+C
=1
3x3ln x−1
9x3+C
Step 3: Therefore, the integral ∫x2ln x dx evaluates to 1
3x3ln x−1
9x3+C,
where Cis the constant of integration.
Question 12
Question
Evaluate the integral ∫x2ln x dx using integration by parts.
Solution
To evaluate the integral ∫x2ln x dx using integration by parts, we need to
choose which part of the integrand to differentiate (letting this be u) and which
part to integrate (letting this be dv).
Let u= ln xand dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Now, we use the formula for integration by parts:
∫u dv =uv −∫v du
Step 1: Compute uv.
ln x·1
3x3=1
3x3ln x
Step 2: Compute ∫v du.
∫1
3x3·1
xdx =1
3∫x2dx =1
3·1
3x3+C=1
9x3+C
Therefore, the integral of ∫x2ln x dx is:
∫x2ln x dx =1
3x3ln x−1
9x3+C
where Cis the constant of integration.
10
Question 13
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we can choose
u= ln(x)and dv =x dx.
Step 1: Compute du and v.
• Let u= ln(x). Then, du =1
xdx.
• Let dv =x dx. Integrating dv gives v=1
2x2.
Step 2: Apply the integration by parts formula ∫u dv =uv −∫v du.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 14
Question
Evaluate the integral ∫xexdx.
Solution
To evaluate the integral ∫xexdx, we will use integration by parts. Integration
by parts is based on the formula ∫u dv =uv −∫v du, where uand vare
differentiable functions of x.
Step 1: Choose uand dv. Let u=xand dv =exdx.
Step 2: Compute du and v. We have du =dx and v=∫exdx =ex.
Step 3: Apply the integration by parts formula.
∫xexdx =uv −∫v du
∫xexdx =x·ex−∫exdx
11
Step 4: Simplify the expression.
∫xexdx =x·ex−ex+C
Therefore, ∫xexdx =xex−ex+C, where Cis the constant of integration.
Question 15
Question
Evaluate the integral ∫x3ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x3ln(x)dx using integration by parts, we will use the
formula ∫u dv =uv −∫v du. Let u= ln(x)and dv =x3dx, then du =1
xdx
and v=1
4x4.
Step 1: Apply integration by parts:
∫x3ln(x)dx =1
4x4ln(x)−∫1
4x4·1
xdx
Step 2: Simplify the integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4∫x3dx
Step 3: Evaluate the remaining integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4·1
4x4+C
Step 4: Simplify the expression:
∫x3ln(x)dx =1
4x4ln(x)−1
16x4+C
Therefore, ∫x3ln(x)dx =1
4x4ln(x)−1
16 x4+C, where Cis the constant of
integration.
Question 16
Question
Evaluate the integral
∫xcos(2x)dx
12
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 we choose uand dv appropriately.
Step 1: Let’s choose u=xand dv = cos(2x)dx. Then, we have:
du =dx and v=1
2sin(2x)
Step 2: Now, applying the integration by parts formula:
∫xcos(2x)dx =uv −∫v du
=x·1
2sin(2x)−∫1
2sin(2x)dx
=1
2xsin(2x)−1
2∫sin(2x)dx
Step 3: Continuing to solve the integral:
∫xcos(2x)dx =1
2xsin(2x)−1
2(−1
2cos(2x)) + C
=1
2xsin(2x) + 1
4cos(2x) + C
Therefore, the solution to the given integral is
∫xcos(2x)dx =1
2xsin(2x) + 1
4cos(2x) + C
where Cis the constant of integration.
Question 17
Question
Evaluate the integral ∫xln x dx using integration by parts.
Solution
To evaluate the integral ∫xln x dx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
13
where uand vare differentiable functions of x.
Step 1: Identify uand dv Let u= ln xand dv =x dx. Then,
du =1
xdx and v=∫x dx =1
2x2
Step 2: Apply integration by parts Using the integration by parts
formula, we have:
∫xln x dx =uv −∫v du = ln x·1
2x2−∫1
2x2·1
xdx
Step 3: Simplify the integral Simplifying the expression gives:
1
2x2ln x−1
2∫x dx
Step 4: Evaluate the integral Now, we can integrate ∫x dx to get:
1
2x2ln x−1
2·1
2x2+C
Step 5: Final answer Therefore, the solution to the integral ∫xln x dx is:
1
2x2ln x−1
4x2+C
where Cis the constant of integration.
Question 18
Question
Evaluate the integral: ∫x2ln(x)dx.
Solution
To evaluate the given integral, we can use integration by parts. Let’s choose
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
Step 2: Simplify the integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3∫x2dx
Step 3: Integrate the remaining term:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
Therefore, the solution to the given integral is 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
14
Question 19
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To solve the integral ∫xln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x dx.
Step 1: Calculate 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
Step 3: Integrate ∫x dx.
∫x dx =1
2x2+C
Step 4: Substitute back into the previous expression.
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 20
Question
Evaluate the integral ∫xarcsin(x)dx.
15
Solution
To evaluate the integral ∫xarcsin(x)dx, we will use integration by parts where
we let u= arcsin(x)and dv =x dx. Then, we will find du and v.
•Step 1: Find du by taking the derivative of u.
u= arcsin(x)
du
dx =1
√1−x2
du =1
√1−x2dx
•Step 2: Find vby integrating dv.
dv =x dx
v=x2
2
•Step 3: Apply integration by parts formula:
∫u dv =uv −∫v du
= arcsin(x)x2
2−∫x2
2·1
√1−x2dx
•Step 4: Integrate ∫x2
2·1
√1−x2dx using the substitution method x=
sin(θ).
∫x2
2·1
√1−x2dx =∫sin2(θ)
2·1
cos(θ)·cos(θ)dθ
=1
2∫sin2(θ)dθ
=1
2∫1−cos(2θ)
2dθ
=1
4(θ−sin(2θ)
2)+C
=1
4(arcsin(x)−x√1−x2)+C
•Step 5: Substitute back xusing the first substitution x= sin(θ)in the
original integral:
∫xarcsin(x)dx = arcsin(x)x2
2−1
4(arcsin(x)−x√1−x2)+C
=1
2x2arcsin(x)−1
4arcsin(x) + 1
4x√1−x2+C
16
Question 21
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral using integration by parts, we need to choose
which part of the integrand to differentiate and which part to integrate. Let’s
follow the integration by parts formula, which states:
∫u dv =uv −∫v du.
Step 1: Choose u= ln(x)and dv =x2dx. Then, calculate du and v:
•du =1
xdx
•v=∫x2dx =x3
3
Step 2: Apply the integration by parts formula:
∫x2ln(x)dx =x3ln(x)
3−∫x3
3·1
xdx.
Step 3: Simplify the integral:
∫x2ln(x)dx =x3ln(x)
3−∫x2
3dx
=x3ln(x)
3−1
3∫x2dx
=x3ln(x)
3−x3
9+C,
where Cis the constant of integration.
Therefore, the integral of x2ln(x)is x3ln(x)
3−x3
9+C.
Question 22
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
17
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x2dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Let’s find du and v.
•u= ln(x)
•du =1
xdx
•dv =x2dx =⇒v=x3
3
Step 2: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫x2ln(x)dx =x3
3ln(x)−∫x3
3·1
xdx
=x3
3ln(x)−1
3∫x2dx
=x3
3ln(x)−1
3·x3
3+C
=x3
3ln(x)−x3
9+C
Therefore, ∫x2ln(x)dx =x3
3ln(x)−x3
9+C, where Cis the constant of
integration.
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 choose:
u= ln(x)
dv =x dx
Step 1: Compute du and v.
du =1
xdx
v=1
2x2
18
Step 2: Apply the integration by parts formula:
∫u dv =uv −∫v du
∫xln(x)dx =1
2x2ln(x)−∫1
2x dx
Step 3: Integrate ∫1
2x dx.
1
2∫x dx =1
2·1
2x2=1
4x2
Step 4: Substitute 1
4x2back into the equation:
1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 24
Question
Evaluate the definite integral:
∫1
0
xsin−1(x)dx
Solution
Step 1: Let u= sin−1(x)and dv =x dx. Then, du =1
√1−x2dx and v=x2
2.
Step 2: Now, we can apply the integration by parts formula:
∫u dv =uv −∫v du
Step 3: Substituting the values of u,v,du, and dv into the formula, we get:
∫1
0
xsin−1(x)dx =[x2
2sin−1(x)]1
0−∫1
0
x2
2·1
√1−x2dx
Step 4: We need to evaluate the integral on the right-hand side. Let’s make
a substitution x= sin(θ), then dx = cos(θ)dθ.
Step 5: After the substitution, the integral becomes:
∫π
2
0
sin2(θ)
2dθ =1
2∫π
2
0
1−cos(2θ)
2dθ
19
Step 6: Now, we can split the integral into two parts and evaluate them
separately:
1
4[θ−sin(2θ)
2]
π
2
0
=π
4
Step 7: Finally, substituting back the values of the definite integral and
simplifying, we get:
∫1
0
xsin−1(x)dx =1
2sin−1(1) −π
4=π
4−π
4= 0
Question 25
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 choose
u= ln(x)and dv =x dx.
Step 1: Calculate du and v.
•u= ln(x)
•du =1
xdx
•v=∫x dx =1
2x2
Step 2: Apply 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
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.
20
Question 2
Question
Evaluate the integral ∫exsin(x)dx using integration by parts.
Solution
To evaluate the integral ∫exsin(x)dx using integration by parts, we will choose
u=exand dv = sin(x)dx. Then, we’ll differentiate uto find du and integrate
dv to find v.
Step 1: Let u=exand dv = sin(x)dx. Then,
du =exdx and v=−cos(x).
Step 2: Now, we can use the integration by parts formula:
∫u dv =uv −∫v du.
Substitute u,v,du, and dv:
∫exsin(x)dx =ex(−cos(x)) −∫(−cos(x))(exdx).
Step 3: Simplify the equation:
=−excos(x) + ∫excos(x)dx.
Step 4: We need to evaluate the new integral ∫excos(x)dx using integra-
tion by parts again. Let u=exand dv = cos(x)dx. Then,
du =exdx and v= sin(x).
Step 5: Apply integration by parts with u=exand dv = cos(x)dx:
∫excos(x)dx =exsin(x)−∫sin(x)exdx.
Step 6: Substitute back the integral into our expression:
=−excos(x)+(exsin(x)−∫exsin(x)dx).
Step 7: Rearrange terms and isolate the original integral on one side:
2∫exsin(x)dx =ex(sin(x)−cos(x)).
Finally, solve for the integral:
∫exsin(x)dx =ex(sin(x)−cos(x))
2+C,
where Cis the constant of integration.
2
Question 3
Question
Evaluate the integral ∫xexsin x dx using integration by parts.
Solution
To evaluate the integral ∫xexsin x dx using integration by parts, we will apply
the formula ∫u dv =uv −∫v du with the appropriate choices for uand dv.
Step 1: Let us choose u=xand dv =exsin x dx. Then, we have:
du =dx and v=∫exsin x dx
Step 2: To find v, we can use integration by parts again by choosing u=
sin xand dv =exdx:
du = cos x dx and v=−excos x+∫excos x dx
Step 3: Now, we need to find ∫excos x dx. Let w=exand dz = cos x dx.
Then, we have:
dw =exdx and z=∫cos x dx = sin x
Step 4: Substituting wand zback, we get:
∫excos x dx =exsin x−∫exsin x dx
Step 5: Now, substituting vand the expression for ∫excos x dx back into
the integration by parts formula, we have:
∫xexsin x dx =uv −∫v du
=x(−excos x)−(−excos x+exsin x−∫exsin x dx)
=−xexcos x+excos x−exsin x+∫exsin x dx
Step 6: Rearranging terms, we get:
2∫xexsin x dx =ex(sin x−xcos x)
∫xexsin x dx =ex(sin x−xcos x)
2+C
So, the solution to the integral ∫xexsin x dx is ex(sin x−xcos x)
2+C, where C
is the constant of integration.
3
Question 4
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 choose
u=xand dv = cos(x)dx. Then, we will find du and vto apply the integration
by parts formula: ∫u dv =uv −∫v du.
Step 1: Find du and v.
u=x, dv = cos(x)dx
du =dx, v =∫cos(x)dx = sin(x)
Step 2: Apply the integration by parts formula.
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C
Therefore, ∫xcos(x)dx =xsin(x) + cos(x) + C, where Cis the constant of
integration.
Question 5
Question
Evaluate ∫xsin(x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(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: Choose uand dv. Let u=xand dv = sin(x)dx. Then, we have:
du =dx and v=−cos(x)
4
Step 2: Apply the integration by parts formula.
∫xsin(x)dx =uv −∫v du
=x(−cos(x)) −∫(−cos(x)dx)
=−xcos(x) + ∫cos(x)dx
Step 3: Integrate ∫cos(x)dx.
∫cos(x)dx = sin(x) + C
Step 4: Substitute back into the equation.
∫xsin(x)dx =−xcos(x) + sin(x) + C
Therefore, ∫xsin(x)dx =−xcos(x) + sin(x) + C, where Cis the constant
of integration.
Question 6
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 follow
the formula:
∫u dv =uv −∫v du
where we choose uand dv appropriately. Let us take u= ln(x)and dv =
x dx.
Step 1: Calculate du and v.
•du =1
xdx
• To find v, we integrate dv:
v=∫x dx =x2
2
5
Step 2: Apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)·x2
2−∫x2
2·1
xdx
=x2ln(x)
2−1
2∫x dx
=x2ln(x)
2−x2
4+C
Therefore, ∫xln(x)dx =x2ln(x)
2−x2
4+C, where Cis the constant of
integration.
Question 7
Question
Evaluate the integral ∫x3ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x3ln(x)dx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let’s choose:
u= ln(x)⇒du =1
xdx
dv =x3dx ⇒v=1
4x4
Step 2: Now, we can apply the integration by parts formula:
∫x3ln(x)dx = ln(x)·1
4x4−∫1
4x4·1
xdx
Step 3: Simplify the integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4∫x3dx
Step 4: Evaluate the remaining integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4·1
4x4+C
6
∫x3ln(x)dx =1
4x4ln(x)−1
16x4+C
Therefore, the integral ∫x3ln(x)dx simplifies to 1
4x4ln(x)−1
16 x4+C, where
Cis the constant of integration.
Question 8
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 choose
u= ln(x)and dv =x2dx.
Step 1: Calculate du and v:
u= ln(x)
dv =x2dx
du =1
xdx
v=1
3x3
Step 2: Apply the formula for integration by parts:
∫x2ln(x)dx =uv −∫v du
=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.
7
Solution
To evaluate ∫xln(x)dx using integration by parts, we will use the formula:
∫u dv =uv −∫v du
Let’s choose u= ln(x)and dv =x dx. Then, we have:
du =1
xdx and v=1
2x2
Step 1: Calculate du and v.
du =1
xdx ⇒du =dx
v=1
2x2
Step 2: Apply integration by parts formula.
∫xln(x)dx =∫u dv
=uv −∫v du
= ln(x)·1
2x2−∫1
2x2dx
=1
2x2ln(x)−1
2∫x2dx
Step 3: Evaluate ∫x2dx.
∫x2dx =1
3x3+C
Step 4: Substitute back into the original integral.
∫xln(x)dx =1
2x2ln(x)−1
2·1
3x3+C
=1
2x2ln(x)−1
6x3+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
6x3+C, where Cis the constant of
integration.
Question 10
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
8
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du.
Step 1: Choose uand dv
Let u= ln(x)and dv =x dx. Then, we have: du =1
xdx and v=1
2x2.
Step 2: Compute du and v
Now, we have: du =1
xdx and v=1
2x2.
Step 3: Apply integration by parts formula
Using the formula ∫u dv =uv −∫v du, we get:
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
Step 4: Simplify the expression
Simplifying, we get: 1
2x2ln(x)−1
2∫x dx
Step 5: Integrate the new integral
Integrating ∫x dx =1
2x2, we have:
1
2x2ln(x)−1
4x2+C
Step 6: Final answer
Therefore, the integral ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the
constant of integration.
Question 11
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 use the
formula ∫u dv =uv −∫v du where uand vare differentiable functions of x.
Step 1: Let’s choose u= ln xand dv =x2dx. Then we have:
du =1
xdx and v=1
3x3
9
Step 2: Now, we apply the integration by parts formula:
∫x2ln x dx =uv −∫v du
= ln x·1
3x3−∫1
3x3·1
xdx
=1
3x3ln x−1
3∫x2dx
=1
3x3ln x−1
3·1
3x3+C
=1
3x3ln x−1
9x3+C
Step 3: Therefore, the integral ∫x2ln x dx evaluates to 1
3x3ln x−1
9x3+C,
where Cis the constant of integration.
Question 12
Question
Evaluate the integral ∫x2ln x dx using integration by parts.
Solution
To evaluate the integral ∫x2ln x dx using integration by parts, we need to
choose which part of the integrand to differentiate (letting this be u) and which
part to integrate (letting this be dv).
Let u= ln xand dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Now, we use the formula for integration by parts:
∫u dv =uv −∫v du
Step 1: Compute uv.
ln x·1
3x3=1
3x3ln x
Step 2: Compute ∫v du.
∫1
3x3·1
xdx =1
3∫x2dx =1
3·1
3x3+C=1
9x3+C
Therefore, the integral of ∫x2ln x dx is:
∫x2ln x dx =1
3x3ln x−1
9x3+C
where Cis the constant of integration.
10
Question 13
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we can choose
u= ln(x)and dv =x dx.
Step 1: Compute du and v.
• Let u= ln(x). Then, du =1
xdx.
• Let dv =x dx. Integrating dv gives v=1
2x2.
Step 2: Apply the integration by parts formula ∫u dv =uv −∫v du.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 14
Question
Evaluate the integral ∫xexdx.
Solution
To evaluate the integral ∫xexdx, we will use integration by parts. Integration
by parts is based on the formula ∫u dv =uv −∫v du, where uand vare
differentiable functions of x.
Step 1: Choose uand dv. Let u=xand dv =exdx.
Step 2: Compute du and v. We have du =dx and v=∫exdx =ex.
Step 3: Apply the integration by parts formula.
∫xexdx =uv −∫v du
∫xexdx =x·ex−∫exdx
11
Step 4: Simplify the expression.
∫xexdx =x·ex−ex+C
Therefore, ∫xexdx =xex−ex+C, where Cis the constant of integration.
Question 15
Question
Evaluate the integral ∫x3ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x3ln(x)dx using integration by parts, we will use the
formula ∫u dv =uv −∫v du. Let u= ln(x)and dv =x3dx, then du =1
xdx
and v=1
4x4.
Step 1: Apply integration by parts:
∫x3ln(x)dx =1
4x4ln(x)−∫1
4x4·1
xdx
Step 2: Simplify the integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4∫x3dx
Step 3: Evaluate the remaining integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4·1
4x4+C
Step 4: Simplify the expression:
∫x3ln(x)dx =1
4x4ln(x)−1
16x4+C
Therefore, ∫x3ln(x)dx =1
4x4ln(x)−1
16 x4+C, where Cis the constant of
integration.
Question 16
Question
Evaluate the integral
∫xcos(2x)dx
12
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 we choose uand dv appropriately.
Step 1: Let’s choose u=xand dv = cos(2x)dx. Then, we have:
du =dx and v=1
2sin(2x)
Step 2: Now, applying the integration by parts formula:
∫xcos(2x)dx =uv −∫v du
=x·1
2sin(2x)−∫1
2sin(2x)dx
=1
2xsin(2x)−1
2∫sin(2x)dx
Step 3: Continuing to solve the integral:
∫xcos(2x)dx =1
2xsin(2x)−1
2(−1
2cos(2x)) + C
=1
2xsin(2x) + 1
4cos(2x) + C
Therefore, the solution to the given integral is
∫xcos(2x)dx =1
2xsin(2x) + 1
4cos(2x) + C
where Cis the constant of integration.
Question 17
Question
Evaluate the integral ∫xln x dx using integration by parts.
Solution
To evaluate the integral ∫xln x dx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
13
where uand vare differentiable functions of x.
Step 1: Identify uand dv Let u= ln xand dv =x dx. Then,
du =1
xdx and v=∫x dx =1
2x2
Step 2: Apply integration by parts Using the integration by parts
formula, we have:
∫xln x dx =uv −∫v du = ln x·1
2x2−∫1
2x2·1
xdx
Step 3: Simplify the integral Simplifying the expression gives:
1
2x2ln x−1
2∫x dx
Step 4: Evaluate the integral Now, we can integrate ∫x dx to get:
1
2x2ln x−1
2·1
2x2+C
Step 5: Final answer Therefore, the solution to the integral ∫xln x dx is:
1
2x2ln x−1
4x2+C
where Cis the constant of integration.
Question 18
Question
Evaluate the integral: ∫x2ln(x)dx.
Solution
To evaluate the given integral, we can use integration by parts. Let’s choose
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
Step 2: Simplify the integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3∫x2dx
Step 3: Integrate the remaining term:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
Therefore, the solution to the given integral is 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
14
Question 19
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To solve the integral ∫xln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x dx.
Step 1: Calculate 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
Step 3: Integrate ∫x dx.
∫x dx =1
2x2+C
Step 4: Substitute back into the previous expression.
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 20
Question
Evaluate the integral ∫xarcsin(x)dx.
15
Solution
To evaluate the integral ∫xarcsin(x)dx, we will use integration by parts where
we let u= arcsin(x)and dv =x dx. Then, we will find du and v.
•Step 1: Find du by taking the derivative of u.
u= arcsin(x)
du
dx =1
√1−x2
du =1
√1−x2dx
•Step 2: Find vby integrating dv.
dv =x dx
v=x2
2
•Step 3: Apply integration by parts formula:
∫u dv =uv −∫v du
= arcsin(x)x2
2−∫x2
2·1
√1−x2dx
•Step 4: Integrate ∫x2
2·1
√1−x2dx using the substitution method x=
sin(θ).
∫x2
2·1
√1−x2dx =∫sin2(θ)
2·1
cos(θ)·cos(θ)dθ
=1
2∫sin2(θ)dθ
=1
2∫1−cos(2θ)
2dθ
=1
4(θ−sin(2θ)
2)+C
=1
4(arcsin(x)−x√1−x2)+C
•Step 5: Substitute back xusing the first substitution x= sin(θ)in the
original integral:
∫xarcsin(x)dx = arcsin(x)x2
2−1
4(arcsin(x)−x√1−x2)+C
=1
2x2arcsin(x)−1
4arcsin(x) + 1
4x√1−x2+C
16
Question 21
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral using integration by parts, we need to choose
which part of the integrand to differentiate and which part to integrate. Let’s
follow the integration by parts formula, which states:
∫u dv =uv −∫v du.
Step 1: Choose u= ln(x)and dv =x2dx. Then, calculate du and v:
•du =1
xdx
•v=∫x2dx =x3
3
Step 2: Apply the integration by parts formula:
∫x2ln(x)dx =x3ln(x)
3−∫x3
3·1
xdx.
Step 3: Simplify the integral:
∫x2ln(x)dx =x3ln(x)
3−∫x2
3dx
=x3ln(x)
3−1
3∫x2dx
=x3ln(x)
3−x3
9+C,
where Cis the constant of integration.
Therefore, the integral of x2ln(x)is x3ln(x)
3−x3
9+C.
Question 22
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
17
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x2dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Let’s find du and v.
•u= ln(x)
•du =1
xdx
•dv =x2dx =⇒v=x3
3
Step 2: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫x2ln(x)dx =x3
3ln(x)−∫x3
3·1
xdx
=x3
3ln(x)−1
3∫x2dx
=x3
3ln(x)−1
3·x3
3+C
=x3
3ln(x)−x3
9+C
Therefore, ∫x2ln(x)dx =x3
3ln(x)−x3
9+C, where Cis the constant of
integration.
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 choose:
u= ln(x)
dv =x dx
Step 1: Compute du and v.
du =1
xdx
v=1
2x2
18
Step 2: Apply the integration by parts formula:
∫u dv =uv −∫v du
∫xln(x)dx =1
2x2ln(x)−∫1
2x dx
Step 3: Integrate ∫1
2x dx.
1
2∫x dx =1
2·1
2x2=1
4x2
Step 4: Substitute 1
4x2back into the equation:
1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 24
Question
Evaluate the definite integral:
∫1
0
xsin−1(x)dx
Solution
Step 1: Let u= sin−1(x)and dv =x dx. Then, du =1
√1−x2dx and v=x2
2.
Step 2: Now, we can apply the integration by parts formula:
∫u dv =uv −∫v du
Step 3: Substituting the values of u,v,du, and dv into the formula, we get:
∫1
0
xsin−1(x)dx =[x2
2sin−1(x)]1
0−∫1
0
x2
2·1
√1−x2dx
Step 4: We need to evaluate the integral on the right-hand side. Let’s make
a substitution x= sin(θ), then dx = cos(θ)dθ.
Step 5: After the substitution, the integral becomes:
∫π
2
0
sin2(θ)
2dθ =1
2∫π
2
0
1−cos(2θ)
2dθ
19
Step 6: Now, we can split the integral into two parts and evaluate them
separately:
1
4[θ−sin(2θ)
2]
π
2
0
=π
4
Step 7: Finally, substituting back the values of the definite integral and
simplifying, we get:
∫1
0
xsin−1(x)dx =1
2sin−1(1) −π
4=π
4−π
4= 0
Question 25
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 choose
u= ln(x)and dv =x dx.
Step 1: Calculate du and v.
•u= ln(x)
•du =1
xdx
•v=∫x dx =1
2x2
Step 2: Apply 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
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.
20
Question 2
Question
Evaluate the integral ∫exsin(x)dx using integration by parts.
Solution
To evaluate the integral ∫exsin(x)dx using integration by parts, we will choose
u=exand dv = sin(x)dx. Then, we’ll differentiate uto find du and integrate
dv to find v.
Step 1: Let u=exand dv = sin(x)dx. Then,
du =exdx and v=−cos(x).
Step 2: Now, we can use the integration by parts formula:
∫u dv =uv −∫v du.
Substitute u,v,du, and dv:
∫exsin(x)dx =ex(−cos(x)) −∫(−cos(x))(exdx).
Step 3: Simplify the equation:
=−excos(x) + ∫excos(x)dx.
Step 4: We need to evaluate the new integral ∫excos(x)dx using integra-
tion by parts again. Let u=exand dv = cos(x)dx. Then,
du =exdx and v= sin(x).
Step 5: Apply integration by parts with u=exand dv = cos(x)dx:
∫excos(x)dx =exsin(x)−∫sin(x)exdx.
Step 6: Substitute back the integral into our expression:
=−excos(x)+(exsin(x)−∫exsin(x)dx).
Step 7: Rearrange terms and isolate the original integral on one side:
2∫exsin(x)dx =ex(sin(x)−cos(x)).
Finally, solve for the integral:
∫exsin(x)dx =ex(sin(x)−cos(x))
2+C,
where Cis the constant of integration.
2
Question 3
Question
Evaluate the integral ∫xexsin x dx using integration by parts.
Solution
To evaluate the integral ∫xexsin x dx using integration by parts, we will apply
the formula ∫u dv =uv −∫v du with the appropriate choices for uand dv.
Step 1: Let us choose u=xand dv =exsin x dx. Then, we have:
du =dx and v=∫exsin x dx
Step 2: To find v, we can use integration by parts again by choosing u=
sin xand dv =exdx:
du = cos x dx and v=−excos x+∫excos x dx
Step 3: Now, we need to find ∫excos x dx. Let w=exand dz = cos x dx.
Then, we have:
dw =exdx and z=∫cos x dx = sin x
Step 4: Substituting wand zback, we get:
∫excos x dx =exsin x−∫exsin x dx
Step 5: Now, substituting vand the expression for ∫excos x dx back into
the integration by parts formula, we have:
∫xexsin x dx =uv −∫v du
=x(−excos x)−(−excos x+exsin x−∫exsin x dx)
=−xexcos x+excos x−exsin x+∫exsin x dx
Step 6: Rearranging terms, we get:
2∫xexsin x dx =ex(sin x−xcos x)
∫xexsin x dx =ex(sin x−xcos x)
2+C
So, the solution to the integral ∫xexsin x dx is ex(sin x−xcos x)
2+C, where C
is the constant of integration.
3
Question 4
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 choose
u=xand dv = cos(x)dx. Then, we will find du and vto apply the integration
by parts formula: ∫u dv =uv −∫v du.
Step 1: Find du and v.
u=x, dv = cos(x)dx
du =dx, v =∫cos(x)dx = sin(x)
Step 2: Apply the integration by parts formula.
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C
Therefore, ∫xcos(x)dx =xsin(x) + cos(x) + C, where Cis the constant of
integration.
Question 5
Question
Evaluate ∫xsin(x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(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: Choose uand dv. Let u=xand dv = sin(x)dx. Then, we have:
du =dx and v=−cos(x)
4
Step 2: Apply the integration by parts formula.
∫xsin(x)dx =uv −∫v du
=x(−cos(x)) −∫(−cos(x)dx)
=−xcos(x) + ∫cos(x)dx
Step 3: Integrate ∫cos(x)dx.
∫cos(x)dx = sin(x) + C
Step 4: Substitute back into the equation.
∫xsin(x)dx =−xcos(x) + sin(x) + C
Therefore, ∫xsin(x)dx =−xcos(x) + sin(x) + C, where Cis the constant
of integration.
Question 6
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 follow
the formula:
∫u dv =uv −∫v du
where we choose uand dv appropriately. Let us take u= ln(x)and dv =
x dx.
Step 1: Calculate du and v.
•du =1
xdx
• To find v, we integrate dv:
v=∫x dx =x2
2
5
Step 2: Apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)·x2
2−∫x2
2·1
xdx
=x2ln(x)
2−1
2∫x dx
=x2ln(x)
2−x2
4+C
Therefore, ∫xln(x)dx =x2ln(x)
2−x2
4+C, where Cis the constant of
integration.
Question 7
Question
Evaluate the integral ∫x3ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x3ln(x)dx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let’s choose:
u= ln(x)⇒du =1
xdx
dv =x3dx ⇒v=1
4x4
Step 2: Now, we can apply the integration by parts formula:
∫x3ln(x)dx = ln(x)·1
4x4−∫1
4x4·1
xdx
Step 3: Simplify the integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4∫x3dx
Step 4: Evaluate the remaining integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4·1
4x4+C
6
∫x3ln(x)dx =1
4x4ln(x)−1
16x4+C
Therefore, the integral ∫x3ln(x)dx simplifies to 1
4x4ln(x)−1
16 x4+C, where
Cis the constant of integration.
Question 8
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 choose
u= ln(x)and dv =x2dx.
Step 1: Calculate du and v:
u= ln(x)
dv =x2dx
du =1
xdx
v=1
3x3
Step 2: Apply the formula for integration by parts:
∫x2ln(x)dx =uv −∫v du
=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.
7
Solution
To evaluate ∫xln(x)dx using integration by parts, we will use the formula:
∫u dv =uv −∫v du
Let’s choose u= ln(x)and dv =x dx. Then, we have:
du =1
xdx and v=1
2x2
Step 1: Calculate du and v.
du =1
xdx ⇒du =dx
v=1
2x2
Step 2: Apply integration by parts formula.
∫xln(x)dx =∫u dv
=uv −∫v du
= ln(x)·1
2x2−∫1
2x2dx
=1
2x2ln(x)−1
2∫x2dx
Step 3: Evaluate ∫x2dx.
∫x2dx =1
3x3+C
Step 4: Substitute back into the original integral.
∫xln(x)dx =1
2x2ln(x)−1
2·1
3x3+C
=1
2x2ln(x)−1
6x3+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
6x3+C, where Cis the constant of
integration.
Question 10
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
8
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du.
Step 1: Choose uand dv
Let u= ln(x)and dv =x dx. Then, we have: du =1
xdx and v=1
2x2.
Step 2: Compute du and v
Now, we have: du =1
xdx and v=1
2x2.
Step 3: Apply integration by parts formula
Using the formula ∫u dv =uv −∫v du, we get:
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
Step 4: Simplify the expression
Simplifying, we get: 1
2x2ln(x)−1
2∫x dx
Step 5: Integrate the new integral
Integrating ∫x dx =1
2x2, we have:
1
2x2ln(x)−1
4x2+C
Step 6: Final answer
Therefore, the integral ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the
constant of integration.
Question 11
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 use the
formula ∫u dv =uv −∫v du where uand vare differentiable functions of x.
Step 1: Let’s choose u= ln xand dv =x2dx. Then we have:
du =1
xdx and v=1
3x3
9
Step 2: Now, we apply the integration by parts formula:
∫x2ln x dx =uv −∫v du
= ln x·1
3x3−∫1
3x3·1
xdx
=1
3x3ln x−1
3∫x2dx
=1
3x3ln x−1
3·1
3x3+C
=1
3x3ln x−1
9x3+C
Step 3: Therefore, the integral ∫x2ln x dx evaluates to 1
3x3ln x−1
9x3+C,
where Cis the constant of integration.
Question 12
Question
Evaluate the integral ∫x2ln x dx using integration by parts.
Solution
To evaluate the integral ∫x2ln x dx using integration by parts, we need to
choose which part of the integrand to differentiate (letting this be u) and which
part to integrate (letting this be dv).
Let u= ln xand dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Now, we use the formula for integration by parts:
∫u dv =uv −∫v du
Step 1: Compute uv.
ln x·1
3x3=1
3x3ln x
Step 2: Compute ∫v du.
∫1
3x3·1
xdx =1
3∫x2dx =1
3·1
3x3+C=1
9x3+C
Therefore, the integral of ∫x2ln x dx is:
∫x2ln x dx =1
3x3ln x−1
9x3+C
where Cis the constant of integration.
10
Question 13
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we can choose
u= ln(x)and dv =x dx.
Step 1: Compute du and v.
• Let u= ln(x). Then, du =1
xdx.
• Let dv =x dx. Integrating dv gives v=1
2x2.
Step 2: Apply the integration by parts formula ∫u dv =uv −∫v du.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 14
Question
Evaluate the integral ∫xexdx.
Solution
To evaluate the integral ∫xexdx, we will use integration by parts. Integration
by parts is based on the formula ∫u dv =uv −∫v du, where uand vare
differentiable functions of x.
Step 1: Choose uand dv. Let u=xand dv =exdx.
Step 2: Compute du and v. We have du =dx and v=∫exdx =ex.
Step 3: Apply the integration by parts formula.
∫xexdx =uv −∫v du
∫xexdx =x·ex−∫exdx
11
Step 4: Simplify the expression.
∫xexdx =x·ex−ex+C
Therefore, ∫xexdx =xex−ex+C, where Cis the constant of integration.
Question 15
Question
Evaluate the integral ∫x3ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x3ln(x)dx using integration by parts, we will use the
formula ∫u dv =uv −∫v du. Let u= ln(x)and dv =x3dx, then du =1
xdx
and v=1
4x4.
Step 1: Apply integration by parts:
∫x3ln(x)dx =1
4x4ln(x)−∫1
4x4·1
xdx
Step 2: Simplify the integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4∫x3dx
Step 3: Evaluate the remaining integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4·1
4x4+C
Step 4: Simplify the expression:
∫x3ln(x)dx =1
4x4ln(x)−1
16x4+C
Therefore, ∫x3ln(x)dx =1
4x4ln(x)−1
16 x4+C, where Cis the constant of
integration.
Question 16
Question
Evaluate the integral
∫xcos(2x)dx
12
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 we choose uand dv appropriately.
Step 1: Let’s choose u=xand dv = cos(2x)dx. Then, we have:
du =dx and v=1
2sin(2x)
Step 2: Now, applying the integration by parts formula:
∫xcos(2x)dx =uv −∫v du
=x·1
2sin(2x)−∫1
2sin(2x)dx
=1
2xsin(2x)−1
2∫sin(2x)dx
Step 3: Continuing to solve the integral:
∫xcos(2x)dx =1
2xsin(2x)−1
2(−1
2cos(2x)) + C
=1
2xsin(2x) + 1
4cos(2x) + C
Therefore, the solution to the given integral is
∫xcos(2x)dx =1
2xsin(2x) + 1
4cos(2x) + C
where Cis the constant of integration.
Question 17
Question
Evaluate the integral ∫xln x dx using integration by parts.
Solution
To evaluate the integral ∫xln x dx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
13
where uand vare differentiable functions of x.
Step 1: Identify uand dv Let u= ln xand dv =x dx. Then,
du =1
xdx and v=∫x dx =1
2x2
Step 2: Apply integration by parts Using the integration by parts
formula, we have:
∫xln x dx =uv −∫v du = ln x·1
2x2−∫1
2x2·1
xdx
Step 3: Simplify the integral Simplifying the expression gives:
1
2x2ln x−1
2∫x dx
Step 4: Evaluate the integral Now, we can integrate ∫x dx to get:
1
2x2ln x−1
2·1
2x2+C
Step 5: Final answer Therefore, the solution to the integral ∫xln x dx is:
1
2x2ln x−1
4x2+C
where Cis the constant of integration.
Question 18
Question
Evaluate the integral: ∫x2ln(x)dx.
Solution
To evaluate the given integral, we can use integration by parts. Let’s choose
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
Step 2: Simplify the integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3∫x2dx
Step 3: Integrate the remaining term:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
Therefore, the solution to the given integral is 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
14
Question 19
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To solve the integral ∫xln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x dx.
Step 1: Calculate 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
Step 3: Integrate ∫x dx.
∫x dx =1
2x2+C
Step 4: Substitute back into the previous expression.
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 20
Question
Evaluate the integral ∫xarcsin(x)dx.
15
Solution
To evaluate the integral ∫xarcsin(x)dx, we will use integration by parts where
we let u= arcsin(x)and dv =x dx. Then, we will find du and v.
•Step 1: Find du by taking the derivative of u.
u= arcsin(x)
du
dx =1
√1−x2
du =1
√1−x2dx
•Step 2: Find vby integrating dv.
dv =x dx
v=x2
2
•Step 3: Apply integration by parts formula:
∫u dv =uv −∫v du
= arcsin(x)x2
2−∫x2
2·1
√1−x2dx
•Step 4: Integrate ∫x2
2·1
√1−x2dx using the substitution method x=
sin(θ).
∫x2
2·1
√1−x2dx =∫sin2(θ)
2·1
cos(θ)·cos(θ)dθ
=1
2∫sin2(θ)dθ
=1
2∫1−cos(2θ)
2dθ
=1
4(θ−sin(2θ)
2)+C
=1
4(arcsin(x)−x√1−x2)+C
•Step 5: Substitute back xusing the first substitution x= sin(θ)in the
original integral:
∫xarcsin(x)dx = arcsin(x)x2
2−1
4(arcsin(x)−x√1−x2)+C
=1
2x2arcsin(x)−1
4arcsin(x) + 1
4x√1−x2+C
16
Question 21
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral using integration by parts, we need to choose
which part of the integrand to differentiate and which part to integrate. Let’s
follow the integration by parts formula, which states:
∫u dv =uv −∫v du.
Step 1: Choose u= ln(x)and dv =x2dx. Then, calculate du and v:
•du =1
xdx
•v=∫x2dx =x3
3
Step 2: Apply the integration by parts formula:
∫x2ln(x)dx =x3ln(x)
3−∫x3
3·1
xdx.
Step 3: Simplify the integral:
∫x2ln(x)dx =x3ln(x)
3−∫x2
3dx
=x3ln(x)
3−1
3∫x2dx
=x3ln(x)
3−x3
9+C,
where Cis the constant of integration.
Therefore, the integral of x2ln(x)is x3ln(x)
3−x3
9+C.
Question 22
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
17
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x2dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Let’s find du and v.
•u= ln(x)
•du =1
xdx
•dv =x2dx =⇒v=x3
3
Step 2: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫x2ln(x)dx =x3
3ln(x)−∫x3
3·1
xdx
=x3
3ln(x)−1
3∫x2dx
=x3
3ln(x)−1
3·x3
3+C
=x3
3ln(x)−x3
9+C
Therefore, ∫x2ln(x)dx =x3
3ln(x)−x3
9+C, where Cis the constant of
integration.
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 choose:
u= ln(x)
dv =x dx
Step 1: Compute du and v.
du =1
xdx
v=1
2x2
18
Step 2: Apply the integration by parts formula:
∫u dv =uv −∫v du
∫xln(x)dx =1
2x2ln(x)−∫1
2x dx
Step 3: Integrate ∫1
2x dx.
1
2∫x dx =1
2·1
2x2=1
4x2
Step 4: Substitute 1
4x2back into the equation:
1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 24
Question
Evaluate the definite integral:
∫1
0
xsin−1(x)dx
Solution
Step 1: Let u= sin−1(x)and dv =x dx. Then, du =1
√1−x2dx and v=x2
2.
Step 2: Now, we can apply the integration by parts formula:
∫u dv =uv −∫v du
Step 3: Substituting the values of u,v,du, and dv into the formula, we get:
∫1
0
xsin−1(x)dx =[x2
2sin−1(x)]1
0−∫1
0
x2
2·1
√1−x2dx
Step 4: We need to evaluate the integral on the right-hand side. Let’s make
a substitution x= sin(θ), then dx = cos(θ)dθ.
Step 5: After the substitution, the integral becomes:
∫π
2
0
sin2(θ)
2dθ =1
2∫π
2
0
1−cos(2θ)
2dθ
19
Step 6: Now, we can split the integral into two parts and evaluate them
separately:
1
4[θ−sin(2θ)
2]
π
2
0
=π
4
Step 7: Finally, substituting back the values of the definite integral and
simplifying, we get:
∫1
0
xsin−1(x)dx =1
2sin−1(1) −π
4=π
4−π
4= 0
Question 25
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 choose
u= ln(x)and dv =x dx.
Step 1: Calculate du and v.
•u= ln(x)
•du =1
xdx
•v=∫x dx =1
2x2
Step 2: Apply 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
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.
20
Question 2
Question
Evaluate the integral ∫exsin(x)dx using integration by parts.
Solution
To evaluate the integral ∫exsin(x)dx using integration by parts, we will choose
u=exand dv = sin(x)dx. Then, we’ll differentiate uto find du and integrate
dv to find v.
Step 1: Let u=exand dv = sin(x)dx. Then,
du =exdx and v=−cos(x).
Step 2: Now, we can use the integration by parts formula:
∫u dv =uv −∫v du.
Substitute u,v,du, and dv:
∫exsin(x)dx =ex(−cos(x)) −∫(−cos(x))(exdx).
Step 3: Simplify the equation:
=−excos(x) + ∫excos(x)dx.
Step 4: We need to evaluate the new integral ∫excos(x)dx using integra-
tion by parts again. Let u=exand dv = cos(x)dx. Then,
du =exdx and v= sin(x).
Step 5: Apply integration by parts with u=exand dv = cos(x)dx:
∫excos(x)dx =exsin(x)−∫sin(x)exdx.
Step 6: Substitute back the integral into our expression:
=−excos(x)+(exsin(x)−∫exsin(x)dx).
Step 7: Rearrange terms and isolate the original integral on one side:
2∫exsin(x)dx =ex(sin(x)−cos(x)).
Finally, solve for the integral:
∫exsin(x)dx =ex(sin(x)−cos(x))
2+C,
where Cis the constant of integration.
2
Question 3
Question
Evaluate the integral ∫xexsin x dx using integration by parts.
Solution
To evaluate the integral ∫xexsin x dx using integration by parts, we will apply
the formula ∫u dv =uv −∫v du with the appropriate choices for uand dv.
Step 1: Let us choose u=xand dv =exsin x dx. Then, we have:
du =dx and v=∫exsin x dx
Step 2: To find v, we can use integration by parts again by choosing u=
sin xand dv =exdx:
du = cos x dx and v=−excos x+∫excos x dx
Step 3: Now, we need to find ∫excos x dx. Let w=exand dz = cos x dx.
Then, we have:
dw =exdx and z=∫cos x dx = sin x
Step 4: Substituting wand zback, we get:
∫excos x dx =exsin x−∫exsin x dx
Step 5: Now, substituting vand the expression for ∫excos x dx back into
the integration by parts formula, we have:
∫xexsin x dx =uv −∫v du
=x(−excos x)−(−excos x+exsin x−∫exsin x dx)
=−xexcos x+excos x−exsin x+∫exsin x dx
Step 6: Rearranging terms, we get:
2∫xexsin x dx =ex(sin x−xcos x)
∫xexsin x dx =ex(sin x−xcos x)
2+C
So, the solution to the integral ∫xexsin x dx is ex(sin x−xcos x)
2+C, where C
is the constant of integration.
3
Question 4
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 choose
u=xand dv = cos(x)dx. Then, we will find du and vto apply the integration
by parts formula: ∫u dv =uv −∫v du.
Step 1: Find du and v.
u=x, dv = cos(x)dx
du =dx, v =∫cos(x)dx = sin(x)
Step 2: Apply the integration by parts formula.
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C
Therefore, ∫xcos(x)dx =xsin(x) + cos(x) + C, where Cis the constant of
integration.
Question 5
Question
Evaluate ∫xsin(x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(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: Choose uand dv. Let u=xand dv = sin(x)dx. Then, we have:
du =dx and v=−cos(x)
4
Step 2: Apply the integration by parts formula.
∫xsin(x)dx =uv −∫v du
=x(−cos(x)) −∫(−cos(x)dx)
=−xcos(x) + ∫cos(x)dx
Step 3: Integrate ∫cos(x)dx.
∫cos(x)dx = sin(x) + C
Step 4: Substitute back into the equation.
∫xsin(x)dx =−xcos(x) + sin(x) + C
Therefore, ∫xsin(x)dx =−xcos(x) + sin(x) + C, where Cis the constant
of integration.
Question 6
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 follow
the formula:
∫u dv =uv −∫v du
where we choose uand dv appropriately. Let us take u= ln(x)and dv =
x dx.
Step 1: Calculate du and v.
•du =1
xdx
• To find v, we integrate dv:
v=∫x dx =x2
2
5
Step 2: Apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)·x2
2−∫x2
2·1
xdx
=x2ln(x)
2−1
2∫x dx
=x2ln(x)
2−x2
4+C
Therefore, ∫xln(x)dx =x2ln(x)
2−x2
4+C, where Cis the constant of
integration.
Question 7
Question
Evaluate the integral ∫x3ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x3ln(x)dx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let’s choose:
u= ln(x)⇒du =1
xdx
dv =x3dx ⇒v=1
4x4
Step 2: Now, we can apply the integration by parts formula:
∫x3ln(x)dx = ln(x)·1
4x4−∫1
4x4·1
xdx
Step 3: Simplify the integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4∫x3dx
Step 4: Evaluate the remaining integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4·1
4x4+C
6
∫x3ln(x)dx =1
4x4ln(x)−1
16x4+C
Therefore, the integral ∫x3ln(x)dx simplifies to 1
4x4ln(x)−1
16 x4+C, where
Cis the constant of integration.
Question 8
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 choose
u= ln(x)and dv =x2dx.
Step 1: Calculate du and v:
u= ln(x)
dv =x2dx
du =1
xdx
v=1
3x3
Step 2: Apply the formula for integration by parts:
∫x2ln(x)dx =uv −∫v du
=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.
7
Solution
To evaluate ∫xln(x)dx using integration by parts, we will use the formula:
∫u dv =uv −∫v du
Let’s choose u= ln(x)and dv =x dx. Then, we have:
du =1
xdx and v=1
2x2
Step 1: Calculate du and v.
du =1
xdx ⇒du =dx
v=1
2x2
Step 2: Apply integration by parts formula.
∫xln(x)dx =∫u dv
=uv −∫v du
= ln(x)·1
2x2−∫1
2x2dx
=1
2x2ln(x)−1
2∫x2dx
Step 3: Evaluate ∫x2dx.
∫x2dx =1
3x3+C
Step 4: Substitute back into the original integral.
∫xln(x)dx =1
2x2ln(x)−1
2·1
3x3+C
=1
2x2ln(x)−1
6x3+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
6x3+C, where Cis the constant of
integration.
Question 10
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
8
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du.
Step 1: Choose uand dv
Let u= ln(x)and dv =x dx. Then, we have: du =1
xdx and v=1
2x2.
Step 2: Compute du and v
Now, we have: du =1
xdx and v=1
2x2.
Step 3: Apply integration by parts formula
Using the formula ∫u dv =uv −∫v du, we get:
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
Step 4: Simplify the expression
Simplifying, we get: 1
2x2ln(x)−1
2∫x dx
Step 5: Integrate the new integral
Integrating ∫x dx =1
2x2, we have:
1
2x2ln(x)−1
4x2+C
Step 6: Final answer
Therefore, the integral ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the
constant of integration.
Question 11
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 use the
formula ∫u dv =uv −∫v du where uand vare differentiable functions of x.
Step 1: Let’s choose u= ln xand dv =x2dx. Then we have:
du =1
xdx and v=1
3x3
9
Step 2: Now, we apply the integration by parts formula:
∫x2ln x dx =uv −∫v du
= ln x·1
3x3−∫1
3x3·1
xdx
=1
3x3ln x−1
3∫x2dx
=1
3x3ln x−1
3·1
3x3+C
=1
3x3ln x−1
9x3+C
Step 3: Therefore, the integral ∫x2ln x dx evaluates to 1
3x3ln x−1
9x3+C,
where Cis the constant of integration.
Question 12
Question
Evaluate the integral ∫x2ln x dx using integration by parts.
Solution
To evaluate the integral ∫x2ln x dx using integration by parts, we need to
choose which part of the integrand to differentiate (letting this be u) and which
part to integrate (letting this be dv).
Let u= ln xand dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Now, we use the formula for integration by parts:
∫u dv =uv −∫v du
Step 1: Compute uv.
ln x·1
3x3=1
3x3ln x
Step 2: Compute ∫v du.
∫1
3x3·1
xdx =1
3∫x2dx =1
3·1
3x3+C=1
9x3+C
Therefore, the integral of ∫x2ln x dx is:
∫x2ln x dx =1
3x3ln x−1
9x3+C
where Cis the constant of integration.
10
Question 13
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we can choose
u= ln(x)and dv =x dx.
Step 1: Compute du and v.
• Let u= ln(x). Then, du =1
xdx.
• Let dv =x dx. Integrating dv gives v=1
2x2.
Step 2: Apply the integration by parts formula ∫u dv =uv −∫v du.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 14
Question
Evaluate the integral ∫xexdx.
Solution
To evaluate the integral ∫xexdx, we will use integration by parts. Integration
by parts is based on the formula ∫u dv =uv −∫v du, where uand vare
differentiable functions of x.
Step 1: Choose uand dv. Let u=xand dv =exdx.
Step 2: Compute du and v. We have du =dx and v=∫exdx =ex.
Step 3: Apply the integration by parts formula.
∫xexdx =uv −∫v du
∫xexdx =x·ex−∫exdx
11
Step 4: Simplify the expression.
∫xexdx =x·ex−ex+C
Therefore, ∫xexdx =xex−ex+C, where Cis the constant of integration.
Question 15
Question
Evaluate the integral ∫x3ln(x)dx using integration by parts.
Solution
To evaluate the integral ∫x3ln(x)dx using integration by parts, we will use the
formula ∫u dv =uv −∫v du. Let u= ln(x)and dv =x3dx, then du =1
xdx
and v=1
4x4.
Step 1: Apply integration by parts:
∫x3ln(x)dx =1
4x4ln(x)−∫1
4x4·1
xdx
Step 2: Simplify the integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4∫x3dx
Step 3: Evaluate the remaining integral:
∫x3ln(x)dx =1
4x4ln(x)−1
4·1
4x4+C
Step 4: Simplify the expression:
∫x3ln(x)dx =1
4x4ln(x)−1
16x4+C
Therefore, ∫x3ln(x)dx =1
4x4ln(x)−1
16 x4+C, where Cis the constant of
integration.
Question 16
Question
Evaluate the integral
∫xcos(2x)dx
12
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 we choose uand dv appropriately.
Step 1: Let’s choose u=xand dv = cos(2x)dx. Then, we have:
du =dx and v=1
2sin(2x)
Step 2: Now, applying the integration by parts formula:
∫xcos(2x)dx =uv −∫v du
=x·1
2sin(2x)−∫1
2sin(2x)dx
=1
2xsin(2x)−1
2∫sin(2x)dx
Step 3: Continuing to solve the integral:
∫xcos(2x)dx =1
2xsin(2x)−1
2(−1
2cos(2x)) + C
=1
2xsin(2x) + 1
4cos(2x) + C
Therefore, the solution to the given integral is
∫xcos(2x)dx =1
2xsin(2x) + 1
4cos(2x) + C
where Cis the constant of integration.
Question 17
Question
Evaluate the integral ∫xln x dx using integration by parts.
Solution
To evaluate the integral ∫xln x dx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
13
where uand vare differentiable functions of x.
Step 1: Identify uand dv Let u= ln xand dv =x dx. Then,
du =1
xdx and v=∫x dx =1
2x2
Step 2: Apply integration by parts Using the integration by parts
formula, we have:
∫xln x dx =uv −∫v du = ln x·1
2x2−∫1
2x2·1
xdx
Step 3: Simplify the integral Simplifying the expression gives:
1
2x2ln x−1
2∫x dx
Step 4: Evaluate the integral Now, we can integrate ∫x dx to get:
1
2x2ln x−1
2·1
2x2+C
Step 5: Final answer Therefore, the solution to the integral ∫xln x dx is:
1
2x2ln x−1
4x2+C
where Cis the constant of integration.
Question 18
Question
Evaluate the integral: ∫x2ln(x)dx.
Solution
To evaluate the given integral, we can use integration by parts. Let’s choose
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
Step 2: Simplify the integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3∫x2dx
Step 3: Integrate the remaining term:
∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C
Therefore, the solution to the given integral is 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
14
Question 19
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To solve the integral ∫xln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x dx.
Step 1: Calculate 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
Step 3: Integrate ∫x dx.
∫x dx =1
2x2+C
Step 4: Substitute back into the previous expression.
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 20
Question
Evaluate the integral ∫xarcsin(x)dx.
15
Solution
To evaluate the integral ∫xarcsin(x)dx, we will use integration by parts where
we let u= arcsin(x)and dv =x dx. Then, we will find du and v.
•Step 1: Find du by taking the derivative of u.
u= arcsin(x)
du
dx =1
√1−x2
du =1
√1−x2dx
•Step 2: Find vby integrating dv.
dv =x dx
v=x2
2
•Step 3: Apply integration by parts formula:
∫u dv =uv −∫v du
= arcsin(x)x2
2−∫x2
2·1
√1−x2dx
•Step 4: Integrate ∫x2
2·1
√1−x2dx using the substitution method x=
sin(θ).
∫x2
2·1
√1−x2dx =∫sin2(θ)
2·1
cos(θ)·cos(θ)dθ
=1
2∫sin2(θ)dθ
=1
2∫1−cos(2θ)
2dθ
=1
4(θ−sin(2θ)
2)+C
=1
4(arcsin(x)−x√1−x2)+C
•Step 5: Substitute back xusing the first substitution x= sin(θ)in the
original integral:
∫xarcsin(x)dx = arcsin(x)x2
2−1
4(arcsin(x)−x√1−x2)+C
=1
2x2arcsin(x)−1
4arcsin(x) + 1
4x√1−x2+C
16
Question 21
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral using integration by parts, we need to choose
which part of the integrand to differentiate and which part to integrate. Let’s
follow the integration by parts formula, which states:
∫u dv =uv −∫v du.
Step 1: Choose u= ln(x)and dv =x2dx. Then, calculate du and v:
•du =1
xdx
•v=∫x2dx =x3
3
Step 2: Apply the integration by parts formula:
∫x2ln(x)dx =x3ln(x)
3−∫x3
3·1
xdx.
Step 3: Simplify the integral:
∫x2ln(x)dx =x3ln(x)
3−∫x2
3dx
=x3ln(x)
3−1
3∫x2dx
=x3ln(x)
3−x3
9+C,
where Cis the constant of integration.
Therefore, the integral of x2ln(x)is x3ln(x)
3−x3
9+C.
Question 22
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
17
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x2dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Let’s find du and v.
•u= ln(x)
•du =1
xdx
•dv =x2dx =⇒v=x3
3
Step 2: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫x2ln(x)dx =x3
3ln(x)−∫x3
3·1
xdx
=x3
3ln(x)−1
3∫x2dx
=x3
3ln(x)−1
3·x3
3+C
=x3
3ln(x)−x3
9+C
Therefore, ∫x2ln(x)dx =x3
3ln(x)−x3
9+C, where Cis the constant of
integration.
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 choose:
u= ln(x)
dv =x dx
Step 1: Compute du and v.
du =1
xdx
v=1
2x2
18
Step 2: Apply the integration by parts formula:
∫u dv =uv −∫v du
∫xln(x)dx =1
2x2ln(x)−∫1
2x dx
Step 3: Integrate ∫1
2x dx.
1
2∫x dx =1
2·1
2x2=1
4x2
Step 4: Substitute 1
4x2back into the equation:
1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 24
Question
Evaluate the definite integral:
∫1
0
xsin−1(x)dx
Solution
Step 1: Let u= sin−1(x)and dv =x dx. Then, du =1
√1−x2dx and v=x2
2.
Step 2: Now, we can apply the integration by parts formula:
∫u dv =uv −∫v du
Step 3: Substituting the values of u,v,du, and dv into the formula, we get:
∫1
0
xsin−1(x)dx =[x2
2sin−1(x)]1
0−∫1
0
x2
2·1
√1−x2dx
Step 4: We need to evaluate the integral on the right-hand side. Let’s make
a substitution x= sin(θ), then dx = cos(θ)dθ.
Step 5: After the substitution, the integral becomes:
∫π
2
0
sin2(θ)
2dθ =1
2∫π
2
0
1−cos(2θ)
2dθ
19
Step 6: Now, we can split the integral into two parts and evaluate them
separately:
1
4[θ−sin(2θ)
2]
π
2
0
=π
4
Step 7: Finally, substituting back the values of the definite integral and
simplifying, we get:
∫1
0
xsin−1(x)dx =1
2sin−1(1) −π
4=π
4−π
4= 0
Question 25
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 choose
u= ln(x)and dv =x dx.
Step 1: Calculate du and v.
•u= ln(x)
•du =1
xdx
•v=∫x dx =1
2x2
Step 2: Apply 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
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.
20
