# Methods of Integration: Integration by Parts ```Calculus II
MAT 146
Methods of Integration: Integration by Parts
Just as the method of substitution is an integration technique that reverses the
derivative process called the chain rule, Integration by parts is a method of
integration that reverses another derivative process, this one called the
product rule.
The Product Rule
For two functions, u(x) and v(x), the product rule for derivatives states that
 d

 d

d
u(x)v(x)   u(x) v(x)  u(x)  v(x).
dx

dx

dx
You may also have seen this written in shorthand form as
d(u  v)  du  v  u  dv.
In words, we might say that “the derivative of the product of two functions is
the derivative of the first function times the second function plus the first
function times the derivative of the second function.”
When we begin with the shorthand representation above, we can create a
formula for integration by parts by integrating each term in the equation:
d(u  v)  du v  u  dv

 d(u  v)   du  v   u  dv
Integrate each term of the equation.

 d(uv)   vdu   udv

uv   vdu   udv

uv   vdu   udv
or
 udv  uv   vdu
Rewrite.
On the left side, the integral of a
derivative gives us the original product.
Subtract  vdu from both sides of the
equation.
Switch left-side and right-side
The last equation in the above derivation is the integration-by-parts formula.
It states that the integral of a product whose factors are a function u and the
derivative of another function v is equivalent to the difference between the
product of the two functions, u and v, and the integral of the product
composed of the function v and the derivative of the function u.
Example 1: Use integration by parts to evaluate  xe x dx ,
where u  x and dv  ex dx .
To apply the integration-by-parts formula, we need to determine du and v.
The table format shown below is a useful way to organize the information:
x
ux
dv  e dx
du  dx
v  ex
We started the table by placing the known information, u and dv, in the top
row. We completed the table by calculating du and v and placing them in the
appropriate columns in the second row. When the table is complete it
x
ux
dv  e dx

du  dx
v  ex

From the table we write the product of u and v () and subtract from it the
integral of the product of v and du (). From the table, we get
x
x
x
 xe dx  xe   e dx
We still are left with an integral expression on the right side of the equation,
but this is one we can evaluate by inspection:
x
x
x
 xe dx  xe   e dx
 xe x  e x  C
Example 2: Use integration by parts to evaluate  x ln xdx .
We have two choices for the substitutions:
u x
u  ln  x 
dv  ln( x)dx
dv  xdx
1
1
dx v  x 2
x
2
Although our choice may not have been immediately clear at the start, by
looking at the two options for substitutions, only one seems promising now.1
We proceed using the information in the right-hand table.
du  dx
v ?
du 
u  ln  x 
du 
dv  xdx

1
1
dx v  x 2
x 
2
1 2 1 2  1 
 x ln  x dx  ln  x   x    x  dx
2  x 
2
1
1
 x 2 ln  x    xdx
2
2
1
1 1 
 x 2 ln  x    x 2   C
2
2 2 
1
1
 x 2 ln  x   x 2  C
2
4
e
Example 3: Use integration by parts to evaluate  ln(x)dx .
1
Determine the substitutions needed for integration by parts:
u  ln  x 
du 
1
1
dx
x
dv  dx
vx
The antiderivative of ln(x) does exist. Derived in Example 3, it is xln(x)-x. For most of us, however, that
antiderivative cannot be determined by inspection, and that’s a goal for integration by parts.
1 
dx
 ln(x)dx  ln(x)  x   x  
x 
1
e
 x  ln(x)   dx
 x  ln(x)  x 1e
 e  ln(e)  e  1 ln(1) 1
 e  e  0 1
1
In these examples, we see the importance of our choices for u and dv. There
are three important guidelines to keep in mind when making these choices.
 When choosing u, our goal should be an expression that simplifies when
it is differentiated. If du is not a simpler expression than u, reconsider
 Your choice for dv should be an expression that can be easily integrated
(see Example 2).
 The integral  vdu should be simpler to evaluate that the original integral
 udv.
The next example illustrates that it could take more than one application of
integration by parts to evaluate an integral.
Example 4: Evaluate  e x sin xdx .
Possible substitutions:
x
ue
dv  sin xdx
du  e x dx
v   cos x
u  sin x
dv  e dx
du  cos xdx
v  ex
x
These are almost identical, except for a negative sign. We’ll proceed here
using the left-hand table to illustrate the importance of sign changes
x
x
x
 e sin xdx  e cosx   e cosxdx
The integral on the right side of the equation is similar to the one we started
with. It’s value is not readily apparent, so we apply the integration-by-parts
technique to it.
For  e x cosxdx , let u = ex and let dv = cosxdx. Then du = exdx and v = sinx.
This gives us  ex cosxdx  ex sin x   ex sin xdx . Substituting the right-hand
expression for the left in the first integration-by-parts application, we get
x
x
x
 e sin xdx  e cosx   e cos xdx
x

 e cosx  e sin x   e sin xdx
x
x
 e x cosx  e x sin x   e x sin xdx
We again seem to be left with an integral that requires further evaluation.
Notice, however, that the integral on the left side of the equation is the same
as the one on the right side. We can simplify the equation above to complete
the evaluation of the original integral:
x
x
x
x
 e sin xdx  e cos x  e sin x   e sin xdx


2  e sin xdx  e cos x  e sin x
x
x

x

1
e x cos x  e x sin x
2
1
 e x sin x  cos x   C
2
You might try evaluating the original integral of Example 4 using the
alternative table of substitutions derived at the beginning of the solution.
What do you find? Is this what you expected?
x
 e sin xdx 
```