Integration by Parts

## **Problem**

**Part 1:** Use the product rule to derive the formula for integration by parts.

**Part 2:** Evaluate the integral

∫
x
(

e
x

ln
⁡

x

)

d
x
.

{\displaystyle \int x(e^x + \ln{x}) \,dx.}

**Part 3:** Evaluate the integral

∫
(
6
−
48

x
2

)

e

−
4

x
2

d
x
.

{\displaystyle \int (6 - 48 x^2) e^{-4x^2} \,dx.}

## **Solution**

**Part 1**

Consider two differentiable functions

u
(
x
)
,

v
(
x
)

{\displaystyle u(x), \, v(x) }

. According to the product rule, we have

d

d
x

(
u
v
# )

d
u

d
x

v
u

d
v

d
x

.

{\displaystyle \frac{d}{dx}(uv) = \frac{du}{dx}v + u\frac{dv}{dx}. }

Integrate both sides of the equation:

∫

d

d
x

(
u
v
)

d
# x
∫

d
u

d
x

v

d
x
∫
u

d
v

d
x

d
x

{\displaystyle \int \frac{d}{dx}(uv) \,dx = \int \frac{du}{dx}v \,dx + \int u\frac{dv}{dx} \,dx }

u
# v
∫
v

d
u

d
x

d
x
∫
u

d
v

d
x

d
x

{\displaystyle uv = \int v\frac{du}{dx} \,dx + \int u\frac{dv}{dx} \,dx }

u
# v
∫
v

d
u
∫
u

d
v
.

{\displaystyle uv = \int v\,du + \int u \,dv. }

Rearrange the equation to obtain the formula pictured above.

∫
u
d
# v
u
v
−
∫
v
d
u
.

{\displaystyle \int udv = uv - \int vdu. }

**Part 2**

Split the integral into the sum of two integrals:

∫

(

x

e
x

x
ln
⁡

x

)

d
# x
∫
x

e
x

d
x
∫
x
ln
⁡

x

d
x
.

{\displaystyle \int \left(xe^x + x\ln{x} \right) dx = \int xe^x \,dx + \int x\ln{x} \,dx.}

For the first integral, let

# u
x

{\displaystyle u = x}

 and

d
# v

e
x

d
x

{\displaystyle dv = e^x dx }

. Therefore,

d
# u
d
x

{\displaystyle du=dx}

 and

# v

e
x

{\displaystyle v = e^x }

. So

u
v
−
∫
v

d
# u
x

e
x

−
∫

e
x

d
x

{\displaystyle uv-\int v \,du = xe^x - \int e^x \,dx }

∫
x

e
x

d
# x
x

e
x

−

e
x

C
1

.

{\displaystyle \int xe^x \,dx = xe^x - e^x + C\_1.}

For the second integral, let

# u
ln
⁡

x

{\displaystyle u = \ln{x} }

 and

d
# v
x
d
x

{\displaystyle dv = x dx }

. Therefore,

d
# u

1
x

{\displaystyle du = \frac{1}{x} }

 and

# v

x
2

2

{\displaystyle v=\frac{x^2}{2}}

. So

u
v
−
∫
v

d
u

=

x
2

2

ln
⁡

x

−
∫

x
2

2

1
x

d
x

=

x
2

2

ln
⁡

x

−
∫

x
2

d
x

{\displaystyle \begin{align}
uv-\int v \,du &= \frac{x^2}{2} \ln{x} - \int \frac{x^2}{2} \frac{1}{x} \,dx \\[5 pt]
&= \frac{x^2}{2} \ln{x} - \int \frac{x}{2} \,dx
\end{align} }

∫
x
ln
⁡

x

d
# x

x
2

2

ln
⁡

x

−

x
2

4

C
2

.

{\displaystyle \int x\ln{x} \,dx = \frac{x^2}{2} \ln{x} - \frac{x^2}{4} + C\_2.}

Combine both results. Let

C

{\displaystyle C}

 be the combined arbitrary constant

# C

C
1

C
2

{\displaystyle C=C\_1+C\_2}

.

∫
x
(

e
x

ln
⁡

x

)

d
# x

e
x

(
x
−
1
)

x
2

2

(

ln
⁡

x

−

1
2

)

C
.

{\displaystyle \int x(e^x + \ln{x}) \,dx = e^x (x-1) + \frac{x^2}{2}\left(\ln{x} - \frac{1}{2}\right) + C.}

**Part 3**

Split the integral into the difference of two integrals

∫
(
6
−
48

x
2

)

e

−
4

x
2

d
# x
6
∫

e

−
4

x
2

d
x
−
48
∫

x
2

e

−
4

x
2

d
x
.

{\displaystyle \int (6 - 48 x^2) e^{-4x^2} \,dx = 6 \int e^{-4x^2} \,dx - 48 \int x^2e^{-4x^2} \,dx.}

For the first integral, let

# u

e

−
4

x
2

{\displaystyle u = e^{-4x^2} }

 and

d
# v
d
x

{\displaystyle dv=dx}

. Therefore,

d
# u
−
8
x

e

−
4

x
2

d
x

{\displaystyle du = -8x e^{-4x^2} \,dx }

 and

# v
x

{\displaystyle v = x }

. So

∫

e

−
4

x
2

d
x

=
x

e

−
4

x
2

−
∫
−
8

x
2

e

−
4

x
2

d
x

=
x

e

−
4

x
2

8
∫

x
2

e

−
4

x
2

d
x
.

{\displaystyle \begin{align}
\int e^{-4x^2} \,dx &= xe^{-4x^2} - \int -8x^2 e^{-4x^2} \,dx \\[5 pt]
&= xe^{-4x^2} + 8 \int x^2 e^{-4x^2} \,dx.
\end{align} }

Therefore,

∫
(
6
−
48

x
2

)

e

−
4

x
2

d
x

=
6

(

x

e

−
4

x
2

8
∫

x
2

e

−
4

x
2

d
x

)

−
48
∫

x
2

e

−
4

x
2

d
x

[
5
p
t
]

=
6
x

e

−
4

x
2

48
∫

x
2

e

−
4

x
2

d
x
−
48
∫

x
2

e

−
4

x
2

d
x

=
6
x

e

−
4

x
2

C
.

{\displaystyle \begin{align}
\int (6 - 48 x^2) e^{-4x^2} \,dx &= 6 \left(xe^{-4x^2} + 8 \int x^2 e^{-4x^2} \,dx \right) - 48 \int x^2e^{-4x^2} \,dx \\ [5 pt]
&= 6 xe^{-4x^2} + 48 \int x^2 e^{-4x^2} \,dx - 48 \int x^2e^{-4x^2} \,dx \\[5 pt]
&= 6xe^{-4x^2} + C.
\end{align}}