Integration by Substitution
So far we have looked at taking integrals of “basic” or
simple terms. What we are going to do next is to take
the integrals of products and quotients that are not as
direct as the problems from section 7.1. Before we do
this we should review the chain rule.
If f (x)  3x 2  5x  , then
4

3
d
 43x 2  5x 6x  5
dx
For this type of problem we took the derivative of the
 exponent) and multiplied it by the
outer function (the
derivative of the inner function, or another way to look
at it is we had f(g(x)) and its derivative if f’(g(x))(g’(x)).
If you recall, the relationship between a function and its
integral is that they are inverses of one another,
meaning that the derivative of the integral gets you back
to the original function. For example;
10
If f (x)  10x 4 than 10 x 4dx  5 x5  c  2 x5  c and if we take the
d
(2x 5  c) 10x 4 , we are back to the original
dx
function. Notice that the  dx just “cancelled out”. (some
derivative:

students find it easier to think of it as the dx becomes the
c) 
For the idea of Substitution we are going to look for a
more complicated function whose derivative is also part of
the function. Substitution is the chain rule in reverse.
Example:
 6x 3x 2  4 dx . This example is a good problem to apply
4
substitution to. It is a fairly cumbersome problem to
2
expand AND it is comprised of a function 3x  4 and its
derivative (6x)


2
All we will do in this type of problem is to replace 3x  4
2
with u and since 6x is the derivative of 3x  4, we can
4
replace the 6x with du. This problem,  6x 3x 2  4 dx can


4
now be simplified to u du

From here we just apply the appropriate rule from 7.1 and
2
substitute 3x  4 back in for all u in the final answer
5
u5
3x 2  4 

 u du  5  c  F(x)  5  c . To check we can take
4
the derivative:
5
4 

2
2
4
d 3x  4  53x  4  
c 
6x   0  3x 2  4  (6x)



dx
5
5



Example 2


2
3
Find  x x  1 dx
In general whenever you have to take the integral of a
 function raised to a power, it is a good time to see if we
can use substitution. For this example we have the terms
x3 and x2, they are not exact derivatives, but they are close.
Since the derivative of x3 is 3x2, if x3 were the u term than
3x2 would be the du. To compensate for the 3, we can
multiply the entire integral by 1/3:
x
2
x
3

 1dx  1  3x 2
3
x
3
1dx

Note that since we are multiplying (1/3)(3) = 1, we have
not changed the value of the integral

3
 12  
1
1 u 2
2 2
du  u   
 c  u  c (substitute x3 + 1 for u)
3    3 3
9
2
3
3
2 3
F(x)  x  12  c
9

Example 3
We can use the same method for some type of rational
expressions as well:

Find
x3
x 2  6x
2
u = x2 + 6x, du = 2x + 6. Similar to the last
example to make the numerator equal to du, we multiply
by 2, to compensate we than multiply the integral by ½



x3
x
2
 6x 
2
=
1
2x  6
2  x2  6x


2

1 du 1 2
  u du 
2

2 u
2
1 u 1
1
1

c
c
c
2
2 1
2u
2 x  6x



Example 4
Just as the rules for
for
e
x
n
are the same as
e
Find
x
were the same as
e
u
n
, the rules
u
2
x

 2x dx :

u will equal x^2 and du will = 2x. Using substitution we
will have


 ex  2x dx 
2
 eudu e u  c  ex  c
2
Example 5


 3  du
2x 
dx
x 2  3x



2
ln
u

c
ln
x
3 c

u 

Example 6
log x
 x dx
K
1
1
u = log x and du = (ln10)x K(du) = x
1
1
 ; k  ln10
(ln10) x x
log x
 x dx 
ln10  udu 
u2
ln10   c 
2
ln10(log x) 2
c
2
Example 7
 p  p  1 dp u = p+ 1 and we can rewrite as p = u – 1
5
5
6
5
(
u

1)
u
du

(
u

u
)du 


7
6
7
6
This gives us u  u  c  ( p  1)  ( p  1)  c
7 6
7
6