Integration by Parts

Introduction

Let's say we want to integrate

$\int x^{2}e^{x^{3}}~dx.$

Here, we can compute this antiderivative by using  $u-$substitution.

While  $u-$substitution is an important integration technique, it will not help us evaluate all integrals.

For example, consider the integral

$\int xe^{x}~dx.$

There is no substitution that will allow us to integrate this integral.

We need another integration technique called integration by parts.

The formula for integration by parts comes from the product rule for derivatives.

Recall from the product rule,

$(f(x)g(x))'=f'(x)g(x)+f(x)g'(x).$

Then, we have

${\begin{array}{rcl}\displaystyle {f(x)g(x)}&=&\displaystyle {\int f'(x)g(x)+f(x)g'(x)~dx}\\&&\\&=&\displaystyle {\int f'(x)g(x)~dx+\int f(x)g'(x)~dx.}\end{array}}$

If we solve the last equation for the second integral, we obtain

$\int f(x)g'(x)~dx=f(x)g(x)-\int f'(x)g(x)~dx.$

This formula is the formula for integration by parts.

But, as it is currently stated, it is long and hard to remember.

So, we make a substitution to obtain a nicer formula.

Let  $u=f(x)$  and  $dv=g'(x)~dx.$

Then,  $du=f'(x)~dx$  and  $v=g(x).$

Plugging these into our formula, we obtain

$\int u~dv=uv-\int v~du.$

Warm-Up

Evaluate the following integrals.

1)   $\int xe^{x}~dx$

Solution:
We have two options when doing integration by parts.
We can let  $u=x$  or  $u=e^{x}.$
In this case, we let  $u$  be the polynomial.
So, we let  $u=x$  and  $dv=e^{x}~dx.$
Then,  $du=dx$  and  $v=e^{x}.$
Hence, by integration by parts, we get
${\begin{array}{rcl}\displaystyle {\int xe^{x}~dx}&=&\displaystyle {xe^{x}-\int e^{x}~dx}\\&&\\&=&\displaystyle {xe^{x}-e^{x}+C.}\end{array}}$

Final Answer:
$xe^{x}-e^{x}+C$

2)   $\int x\cos(2x)~dx$

Solution:
We have a choice to make.
We can let  $u=x$  or  $u=\cos(2x).$
In this case, we let  $u$  be the polynomial.
So, we let  $u=x$  and  $dv=\cos(2x)~dx.$
Then,  $du=dx.$
By  $u-$substitution,  $v=\int \cos(2x)~dx={\frac {1}{2}}\sin(2x).$
Hence, by integration by parts, we get
${\begin{array}{rcl}\displaystyle {\int x\cos(2x)~dx}&=&\displaystyle {{\frac {1}{2}}x\sin(2x)-\int {\frac {1}{2}}\sin(2x)~dx}\\&&\\&=&\displaystyle {{\frac {1}{2}}x\sin(2x)+{\frac {1}{4}}\cos(2x)+C,}\end{array}}$
where we use  $u-$ substitution to evaluate the last integral.

Final Answer:
${\frac {1}{2}}x\sin(2x)+{\frac {1}{4}}\cos(2x)+C$

3)   $\int \ln x~dx$

Solution:
We have a choice to make.
We can let  $u=\ln x$  or  $dv=\ln x~dx.$
In this case, we don't want to let  $dv=\ln x~dx$  since we don't know how to integrate  $\ln x$  yet.
So, we let  $u=\ln x$  and  $dv=1~dx.$
Then,  $du={\frac {1}{x}}~dx$  and  $v=x.$
Hence, by integration by parts, we get
${\begin{array}{rcl}\displaystyle {\int \ln x~dx}&=&\displaystyle {x\ln(x)-\int x{\big (}{\frac {1}{x}}{\big )}~dx}\\&&\\&=&\displaystyle {x\ln(x)-\int 1~dx}\\&&\\&=&\displaystyle {x\ln(x)-x+C.}\end{array}}$

Final Answer:
$x\ln(x)-x+C$

Exercise 1

Evaluate  $\int x\sec ^{2}x~dx.$

Since we know the antiderivative of  $\sec ^{2}x,$

we let  $u=x$  and  $dv=\sec ^{2}x~dx.$

Then,  $du=dx$  and  $v=\tan x.$

Using integration by parts, we get

${\begin{array}{rcl}\displaystyle {\int x\sec ^{2}x~dx}&=&\displaystyle {x\tan x-\int \tan x~dx}\\&&\\&=&\displaystyle {x\tan x-\int {\frac {\sin x}{\cos x}}~dx.}\end{array}}$

For the remaining integral, we use  $u-$substitution.

Let  $u=\cos x.$  Then,  $du=-\sin x~dx.$

So, we get

${\begin{array}{rcl}\displaystyle {\int x\sec ^{2}x~dx}&=&\displaystyle {x\tan x+\int {\frac {1}{u}}~du}\\&&\\&=&\displaystyle {x\tan x+\ln |u|+C}\\&&\\&=&\displaystyle {x\tan x+\ln |\cos x|+C.}\end{array}}$

So, we have

$\int x\sec ^{2}x~dx=x\tan x+\ln |\cos x|+C.$

Exercise 2

Evaluate  $\int {\frac {\ln x}{x^{3}}}~dx.$

We start by letting  $u=\ln x$  and  $dv={\frac {1}{x^{3}}}~dx.$

Then,  $du={\frac {1}{x}}~dx$  and  $v={\frac {-1}{2x^{2}}}.$

So, using integration by parts, we get

${\begin{array}{rcl}\displaystyle {\int {\frac {\ln x}{x^{3}}}~dx}&=&\displaystyle {{\frac {-\ln x}{2x^{2}}}-\int {\frac {-1}{2x^{2}}}{\big (}{\frac {1}{x}}{\big )}~dx}\\&&\\&=&\displaystyle {{\frac {-\ln x}{2x^{2}}}+\int {\frac {1}{2x^{3}}}~dx}\\&&\\&=&\displaystyle {{\frac {-\ln x}{2x^{2}}}-{\frac {1}{4x^{2}}}+C.}\end{array}}$

So, we have

$\int {\frac {\ln x}{x^{3}}}~dx={\frac {-\ln x}{2x^{2}}}-{\frac {1}{4x^{2}}}+C.$

Exercise 3

Evaluate  $\int x{\sqrt {x+1}}~dx.$

We start by letting  $u=x$  and  $dv={\sqrt {x+1}}~dx.$

Then,  $du=dx.$

Also, by  $u-$substitution, we have

$v=\int {\sqrt {x+1}}~dx={\frac {2}{3}}(x+1)^{\frac {3}{2}}.$

Hence, by integration by parts, we get

${\begin{array}{rcl}\displaystyle {\int x{\sqrt {x+1}}~dx}&=&\displaystyle {{\frac {2}{3}}x(x+1)^{\frac {3}{2}}-\int {\frac {2}{3}}(x+1)^{\frac {3}{2}}~dx}\\&&\\&=&\displaystyle {{\frac {2}{3}}x(x+1)^{\frac {3}{2}}-{\frac {4}{15}}(x+1)^{\frac {5}{2}}+C,}\end{array}}$

where we use  $u-$ substitution to evaluate the last integral.

So, we have

$\int x{\sqrt {x+1}}~dx={\frac {2}{3}}x(x+1)^{\frac {3}{2}}-{\frac {4}{15}}(x+1)^{\frac {5}{2}}+C.$

Exercise 4

Evaluate  $\int x^{2}e^{-2x}~dx.$

We start by letting  $u=x^{2}$  and  $dv=e^{-2x}~dx.$

Then,  $du=2x~dx.$

Also, by  $u-$substitution, we have

$v=\int e^{-2x}~dx={\frac {e^{-2x}}{-2}}.$

Hence, by integration by parts, we get

${\begin{array}{rcl}\displaystyle {\int x^{2}e^{-2x}~dx}&=&\displaystyle {{\frac {x^{2}e^{-2x}}{-2}}-\int 2x{\frac {e^{-2x}}{-2}}~dx}\\&&\\&=&\displaystyle {{\frac {x^{2}e^{-2x}}{-2}}+\int xe^{-2x}~dx.}\end{array}}$

Now, we need to use integration by parts a second time.

Let  $u=x$  and  $dv=e^{-2x}~dx.$

Then,  $du=dx$  and  $v={\frac {e^{-2x}}{-2}}.$

Therefore, using integration by parts again, we get

${\begin{array}{rcl}\displaystyle {\int x^{2}e^{-2x}~dx}&=&\displaystyle {{\frac {x^{2}e^{-2x}}{-2}}+{\frac {xe^{-2x}}{-2}}-\int {\frac {e^{-2x}}{-2}}~dx}\\&&\\&=&\displaystyle {{\frac {x^{2}e^{-2x}}{-2}}+{\frac {xe^{-2x}}{-2}}+{\frac {e^{-2x}}{-4}}+C.}\end{array}}$

So, we have

$\int x^{2}e^{-2x}~dx={\frac {x^{2}e^{-2x}}{-2}}+{\frac {xe^{-2x}}{-2}}+{\frac {e^{-2x}}{-4}}+C.$

Exercise 5

Evaluate  $\int {\frac {\ln x}{x^{3}}}~dx.$

First, using the Quotient Rule, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin x(e^{x})'-e^{x}(x^{2}\sin x)'}{(x^{2}\sin x)^{2}}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\sin x)'}{x^{4}\sin ^{2}x}}.}\end{array}}$

Now, we need to use the Product Rule. So, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}(\sin x)'+(x^{2})'\sin x)}{x^{4}\sin ^{2}x}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.}\end{array}}$

So, we have

$f'(x)={\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.$

Exercise 6

Evaluate  $\int \sin(2x)\cos(3x)~dx.$

First, using the Quotient Rule, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin x(e^{x})'-e^{x}(x^{2}\sin x)'}{(x^{2}\sin x)^{2}}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\sin x)'}{x^{4}\sin ^{2}x}}.}\end{array}}$

Now, we need to use the Product Rule. So, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}(\sin x)'+(x^{2})'\sin x)}{x^{4}\sin ^{2}x}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.}\end{array}}$

So, we have

$f'(x)={\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.$