Section 7.2B Notes Page 1
7.2B Natural Logarithms: Integration
x n 1
as long as n is not equal to -1. Now we
n 1
1
will look at what happens when n is -1. We are looking at this case:  dx . The following is how we
x
integrate this:
Previously we looked at the power rule for integration:
n
 x dx 
Integration of a Natural Logarithm
Let u be a differentiable function of x. Then:
1
1.)
x
2.)
1
 u du  ln u  C Since du  u  dx we can rewrite this as:
dx  ln x  C
u
 u du  ln u  C
We have the absolute value symbols here so that any x value will fit the domain of the natural logarithm.
EXAMPLE: Find the indefinite integral:
1
 x5
dx .
We will be using substitution for most of the problems in this section. Here u  x  5 . Then du  dx . Now
1
we make our substitution:  du . When we integrate this we get ln u  C . Then we can replace the u with
u
x  5 and we get: ln x  5  C as our answer.
EXAMPLE: Find the indefinite integral:
x2
 3  x 3 dx .
Using substitution we will get u  3  x 3 . Then du  3 x 2 dx . Solving for dx we get: dx 
du
. Now we
 3x 2
1 1
x 2 du
. This simplifies to:   du . When we integrate this we get
make our substitution:  
2
u  3x
3 u
1
 ln u  C . Then we can replace the u with 3  x 3 and we get:
3
1
 ln 3  x 3  C which is our answer.
3
Section 7.2B Notes Page 2
EXAMPLE: Find the indefinite integral:

x
9  x2
dx .
Using substitution we get u  9  x 2 . Then du  2 x dx . Solving for dx we get: dx 
du
. Now we make
 2x
1
1 1
du
1 
 u  2 x . This simplifies to:  2  u du . This can be rewritten as:  2  u 2 du Is a
natural log needed on this one? The answer is no because the power on the u is not -1. When we integrate this
1
1
 u2
1
 C . This simplifies to:  u 2  C . Then we can replace the u with 9  x 2 and we get:
we get 2
1
2
2
 9  x  C which is our answer.
our substitution:
x

EXAMPLE: Find the indefinite integral:
x
3
x( x  2)
dx .
 3x 2  4
Using substitution we will get u  x 3  3x 2  4 . Then du  3x 2  6 x dx . Solving for dx and after factoring we
du
x( x  2)
du
1 1
. Now we make our substitution: 
. This simplifies to:  du .

get: dx 
3 u
3x( x  2)
u
3 x( x  2)
1
When we integrate this we get ln u  C . Then we can replace the u with x 3  3 x 2  4 and we get:
3
1
ln x 3  3x 2  4  C which is our answer.
3
EXAMPLE: Find the indefinite integral:
2x 2  7 x  3
 x  2 dx .
The problem with this one is that if we use the numerator for u and then take the derivative it won’t cancel out
the denominator. Therefore we need to use long division to break this up into separate expressions:
2 x  11
x  2 2x  7 x  3
First we ask ourselves how many times x goes into 2x 2 . We ignore the -2 at the
2x 2  4x
11x  3
11x  22
19
moment. We know x goes into 2x 2 an amount of 2x. We then multiply x – 2 by
2x to get 2 x 2  4 x . We write this on the next line and then we subtract. Then
we get 11x  3 . We now ask how many times does x go into 11x. The answer is
11. We then multiply x – 2 by 11 to get 11x  22 . We subtract and get a
remainder of 19.
2
Section 7.2B Notes Page 3
2x  7 x  3
19
dx =  2 x  11 
dx . Now we can integrate each part separately.
x2
x2
19
du . Integrating
For the last part, we will let u  x  2 . Then du  dx . Making the substitution we get: 
u
we get 19 ln u  C . Putting this all together and integrating the first two terms we will get:
So now we know that
2

x 2  11x  19 ln x  2  C as our answer.
EXAMPLE: Find the indefinite integral:
3x 3  x 2  x  2
dx .
 x2  2
The problem with this one is that if we use the numerator for u and then take the derivative it won’t cancel out
the denominator. Therefore we need to use long division to break this up into separate expressions. Remember
that all of the terms must be ordered in descending powers and if there is a missing term you must put in a zero
place keeper.
3x  1
x  0 x  2 3x  x  x  2
2
3
2
Remember you are always subtracting when doing long division.
3x 2  0 x 2  6 x
 x 2  5x  2
 x 2  0x  2
 5x
3x 3  x 2  x  2
5x
dx =  3x  1  2
dx . Now we can integrate each
 x2  2
x 2
du
part separately. For the last part, we will let u  x 2  2 dx . So du  2 x dx . Solving for dx you get: dx 
.
2x
5 1
5 x du
. Simplifying you will get:  du . Integrating we get
Making the substitution we get:  
2 u
u 2x
5
3
5
ln u  C . Putting this all together and integrating the first two terms we will get: x 2  x  ln x 2  2  C .
2
2
2
So our problem can be rewritten as:
EXAMPLE: Find the indefinite integral:

1
1


x 1  x 3 


2
3
2
1
Using substitution we will get u  1  x 3 . Then du 
make our substitution:

1
2
3
x u
dx .
2
 3x 3 du . This simplifies to: 3
1
2
1 3
x dx . Solving for dx we get: dx  3 x 3 du . Now we
3
1
du . When we integrate this we get
u
1
3 ln u  C . Then we can replace the u with 1  x 3 and we get: 3 ln 1  x 3  C as our answer.
Section 7.2B Notes Page 4
EXAMPLE: Find the indefinite integral:
1
1
3x
dx .
1
1
Using substitution we will get u  1  (3 x) 2 . Then du 
you will get: du 
3
2 3x

1
(3 x) 2 (3) dx by using the chain rule. Simplifying
2
dx . Solving for dx we get: dx 
2 3x
du . Now we make our substitution:
3
1 2 3x
2
3x
du . This simplifies to: 
du . We need to make another substitution to get rid of the
3
3 u
square root. We will take our original definition for u and solve for 3 x to get: 3 x  u  1 . Now we can
u
1
2 u 1
2
du . This simplifies to:  1  du . When we integrate this we get

3 u
3
u
2
2
2
2
u  ln u  C . Then we can replace the u with 1  3 x and we get:
1  3x  ln 1  3 x  C .
3
3
3
3
make another substitution:

EXAMPLE: Find the indefinite integral:

 tan x dx .
We haven’t done this one yet. First let’s use identities to write this as:
sin x
 cos x
dx . Now we can use
substitution to integrate this. We will let u  cos x . Then du   sin x dx . Solving for dx you will get:
1
du
sin x
du

. Simplifying will give us:   du . Integrating
dx  
. Now we make our substitution: 
sin x
u
u
sin x
this will give:  ln u  C . Then we can replace the u with cos x and we get:  ln cos x  C .
We can do a similar process for cot x , sec x , and csc x to get the following results:
Integrals of the Six Trigonometric Functions
 sin u
du   cos u  C
 cos u
du  sin u  C
 tan u
du   ln cos u  C
 cot u
du  ln sin u  C
 sec u
du  ln sec u  tan u  C
 csc u
du   ln csc u  cot u  C
More on next page…
Section 7.2B Notes Page 5
x
 sec 2
EXAMPLE: Find the indefinite integral:
Using substitution we will get u 
substitution:
 sec u  2du .
dx .
1
x
. Then du 
dx . Solving for dx we get: dx  2du . Now we make our
2
2
This simplifies to: 2  sec u du . When we integrate this we get
2 ln sec u  tan u  C . Then we can replace the u with
x
x
x
and we get: 2 ln sec  tan  C .
2
2
2
 csc  cot 
EXAMPLE: Find the indefinite integral:
d .
We can integrate each thing separately by using the integration formulas:  ln csc   cot   ln sin x  C
We can use log properties to write this as: ln
sin 
 C . This can also be simplified by using identities:
csc   cot 
sin 2 
sin 
sin 
 C . This equals: ln
 C which simplifies to: ln
 C . We can use the
ln
1
cos 
1  cos 
1  cos 

sin  sin 
sin 
identity sin 2   1  cos 2  : ln
1  cos 1  cos   C
1  cos 2 
 C . Now we can factor the numerator: ln
1  cos 
1  cos 
This simplifies to: ln 1  cos   C .
e2
EXAMPLE: Find the definite integral:
1
 x ln x
dx .
e
1
dx . Solving for dx we get: dx  x du . Now we make
x
Using substitution we will let u  ln x . Then du 
e2
1
 x du . This simplifies to:
our substitution: 
x u
e
can replace the u with ln x and we get:
ln ln x 
e2
e
e2
1
u
du . When we integrate this we get ln u . Then we
e
. Now we plug in our top limit and our bottom limit:
ln ln e 2  ln ln e . This can be rewritten as: ln 2 ln e  ln ln e . We also know that ln e  1 , so our problem
becomes: ln 2  ln 1 . Since ln 1  0 our answer is ln 2 .
EXAMPLE: Find the indefinite integral:
5
 (5x  4)  ln(5x  4)
Using substitution we will let u  ln(5 x  4) . Then du 
substitution:
5
 (5x  4)  u 
(5 x  4)
du . This gives
5
1
u
dx .
5
(5 x  4)
dx , so dx 
du . Now we make our
5x  4
5
du  ln u  c  lnln(5 x  4)   C .