Partial Fractions
& Integration
1.Introduction
To evaluate
5
 x  2x  3 dx
we need to be able to split the fraction into what are
called Partial Fractions.
Adding fractions
1

 x  2
1

 x  3
 x  3   x  2 
 x  2  x  3

5
 x  2  x  3
This implies that in doing the reverse process
5
“Express
in partial fractions” we will get 2 fractions of the form
x  2x  3
B
A
and
x  2
x  3
Example Express
5
x  2x  3
in partial fractions
5

A
B

x2 x3
Let
x  2x  3
gives
A( x  3)  B( x  2)  5
The values of x  2 and x  3 makes one bracket zero giving the value of one
unknown in each case.
A( 2  3)  B( 2  2)  5
Putting x  2
A 1
A( 3  3)  B( 3  2)  5
Putting x  3
B  1
5
1
1
So


x  2x  3 x  2 x  3
Which is where we started!
Exercise 1
Simplify the following ie express them as single fractions
1.
1

 x  1
3.
1

x 1
2
 2 x  1
1
 x  1
2

2.
1
 x  1
3
4.
2x
3
2x  1
x 1
3
5

2
 x  1  x  3
2

2. Some Partial Fractions
(i) Proper fractions
x 2  9 x  25
is a proper fraction as the highest power in the numerator is 2.
x  12 x  1x  5
x3
is improper as the highest power in both numerator and
x  12 x  1x  5
denominator is 3.
x 4  3x 2  x  1
is improper as the highest power in numerator (4) is greater than
x  12 x  1x  5
the highest power in the denominator (3).
In the following they are all proper fractions.
Type 1: denominator with only linear factors such as question 1 above
f ( x)
A
B
C



x  12 x  1x  5 x  1 2 x  1 x  5
Type 2: denominator with a quadratic factor such as question 2 above
f ( x)
Ax  B
C
D



x 2  1 2 x  1x  5 x 2  1 2 x  1 x  5


Type 3: denominator with repeated linear factor such as question 3 above
f ( x)
A
B
C
D




3
3
2 x  1 x  5
x  1 2 x  1 x  1 x  1
(a) - Type 1
Express
3x  1
in partial fractions
x  12 x  1
3x  1
A
B


Let
x  12 x  1 x  1 2 x  1
gives
put x  1
put x  
Solution
1
2
3x  1  A  2 x  1  B  x  1
3  1  A  2  1
 A  43
3   12   1  B   12  1
 B  13
3x  1
4
1


x  12 x  1 3x  1 32 x  1
1
(b) - Type 2
Express
3x  1
x  1x 2  1
in partial fractions
3x  1
x  1x2  1
Let


0  A  B  B  2
1 AC  C 1
RHS =  B + C  (2)  1  3  LHS)
equate coefficient of x 2
equate constant term
(check coeff of x
3x  1
A
Bx  C

x  1 x2  1
3x  1  Ax 2  1  Bx  C x  1
3  1  A1  1  A  2
putting x  1
Solution

x  1x2  1

2
 2x  1
2
2x  1



x  1 x2  1
x  1 x2  1




(c) - Type 3
Express
36
x  12 x  5
in partial fractions
36
Let

A
x  12 x  5 x  12

B
C

x  1 x  5
Multiplying through by x  12 x  5
36  Ax  5  Bx  5x  1  C x  12
36  A1  5
 A6
gives
putting x  1
putting x  5
equate coefficient of x 2
(check put x  0
Solution
36
36  C  1  52  C  1
0  BC
 B  1
RHS = 5 A  5B  C  30  5(1)  1  36  LHS)

6
x  12 x  5 x  12

1
1

x  1 x  5
2
(ii) Improper Fractions
x 4  2 x3  x 2  4 x  4
x  3x 2  1
The fraction
is an improper fraction as the highest power of the
numerator is 4 and of the denominator is 3.
Before expressing it in partial fractions it is necessary to divide through.
This gives
x 4  2 x3  x 2  4 x  4
x  3x 2  1
x2  2 x  7
x  3x2  1
The fraction
x  3x
2

1
x2  2x  7
x  3x 2  1
can then be put into partial fractions as in example 3 above
x 4  2 x3  x 2  4 x  4
giving
 x 1
 x 1
1
2

x  3 x2  1


Exercise 2
Express the following in Partial Fractions
1.
4.
7.
10.
3x
x  2x  1

4
x x2  4

2.
5.
2
x  1 x  1
2
x4  1
3
x  2x
8.
2x  1
x  1x  2x  3
3x 2  2 x
3.
x  2x 2  4
6.
x 2  3x
9.
x2  4
1
11.
x x  1 x  1
12.
2x
x 2  25
x2  1


x x2  1
5x  3
x  2x  32
16
x  12 x  13
3
3. Integration
Note It is important to recognise certain standard integrals and methods here.
f ' ( x)
dx  ln f ( x )  c
1. The use of 
f ( x)
2. Using constant multipliers to simplify integration such as
x
1
4x
1
dx  
dx  ln 2 x 2  3  c
 2
2
4 2x  3
4
2x  3


3. Splitting up such expressions as

giving
2x  1
x 1
Examples
(a) Simplify
dx  
2
2x
2
x 1
 3  x  1

1
dx
3  2 x  1

4
3

1
6

1
ln 2 x  1  c
6
4
1
  x  1 dx
4
ln x  1
3
2
  2 x  1

2
 x 1
2
 x 1
x 1


2x
2
x 1

1
2
x 1
dx  ln x 2  1  tan 1 x  c
3x  1
  x  1 2 x  1 dx
dx
3x  1
 x  1x 2  1 dx
From example. (b) on page 2

x 1
to get
3x  1

(b) Simplify
1
2
2
 x  12x  1 dx
From example. (a) on page 1


2x  1


1  2x
x2  1
1
2
x 1
3x  1
  x  1


x2  1
dx
dx

2x
2
x 1
dx


 2 ln  x  1  tan1 x  ln x 2  1  c
4
(c) Find the value of
4
 2 x  12 x  5dx
36
From example (c) on page 2 we can write
4
4
36
6
1
dx 


2 x 1 2 x  5
2 x 1 2
x

1


  

 


1
dx
 x  5
4

 6

 ln x  1  ln x  5 
 x 1
2


 6
 6
   ln 3  ln 9     ln1  ln 7 
3
1


 2  ln 3  ln 9  6  0  ln 7
 97 
 3
 4  ln 
 4  ln  

 3 
7
5
(d) Evaluate 
4
24 x3  x  3
 x  1 2 x  1
dx
Multiplying out top and bottom and dividing gives:
24 x3  x  3
 x  1 2 x  1

24 x 4  72 x3
2 x2  x  1
 12 x 2  30 x  9 
Expressing
39 x  9
2
2x  x  1
Let

39 x  9
2 x2  x  1
39 x  9
in partial fractions
 x  1 2 x  1
39 x  9
A
B


 x  1 2 x  1 x  1 2 x  1
39 x  9  A  2 x  1  B  x  1
then
putting x  1 gives 48  3 A  A  16
putting x  0 gives
This gives
Hence
9  A B  B  7
24 x3  x  3
 x  1 2 x  1

5 24 x3 x  3

 12 x 2  30 x  9 
5
 4  x  1 2 x  1 dx   4 12 x
2
16
7

 x  1  2 x  1
 30 x  9 
16
7

dx
 x  1  2 x  1
5
  4 x3  15 x 2  9 x  16 ln x  1  7 ln 2 x  1 
2

4
5

 
 500  375  45  16 ln 4  7 ln11  256  240  36  16 ln 3  7 ln 9
2
2
 80  32 ln 2  7 ln11   20  16 ln 3  7 ln 3

2
7
 80  32 ln 2  ln11  20  16 ln 3  7 ln 3
2
 100  32 ln 2  23 ln 3  7 ln11
2
(e) Evaluate
First express
1 x
2
1

1 x




x2 x2  1

A
x2
dx
in Partial Fractions
x2 x2  1
1 x
Writing

x2 x2  1

B Cx  D

x
x2  1
and equating terms gives A  1, B  1, C  1 , D  1
1 x
2
1
Hence


x2 x2  1
dx 

2
1
1
2
1
x2

x 2 
1 x 1

dx
x x2  1
1 1  2x 
1
 
dx
 2
2
x 2  x 1 x 1


2
   x 1  ln x  1 ln x 2  1  tan1 x 
2

 1

 

  1  ln 2  1 ln 5  tan1 2  1  ln1  1 ln 2  tan1 1
2
2
2
This can be simplified depending on what is required.
Exercise 3
Find the following integrals (You will find your solutions to the previous exercise will
help in some cases)
5
10(2 x  1)
3x
2.
dx
1.
dx
x  1x  2x  3
2  x  2  x  1


3.

2x
2
x  25
2
4.
dx
3x 2  2 x
5.
0
7.
 x  12 x  1 dx
9.
x  2x 2  4
dx
6.
4
8.
5x  3
 x  2x  32 dx
10.
2
1 xx2  4 dx
4
x2  1
 xx2  1 dx



8 2 x 2  3x
dx
2
6
x 4
4
2 x 1
dx
1 3

x  2x
6
Answers
Exercise 1
3
1.
 x  1 2 x  1
3.
2.
x2  x  1
4.
 x  1
3
7 x2  2 x  3
 x2  1  2x  1
2  x2  4
 x  1 x  3
Exercise 2
2
1
1.

x  2 x 1
1
1
3.

x5 x5
1
2x  2
5.

x  2 x2  4
1
1
1
7.


2 2 x  1 2 x  1
x  1
9.
11.
12
x  32

2.
4.
6.
8.
7
7

x3 x2
1
1
1


2x  1 2x  1 x
10.
12.
2
x  12

1
1
7


2 x  1 5 x  2  10 x  3
1
x

2
x x 4
1
1
1


x 1 x 1 x
5
1
1

2 x  2  2 x  2 
1
5x
x

2 x 2 x2  2


3
4
4
3



x  1 x  13 x  12 x  1
Exercise 3
Note: most of the partial fractions were worked out in the previous exercise.
The „best‟ form of the answer depends on what comes next!
1. 2 ln 7
2
2. 5 ln x  1  2 ln x  2  7 ln x  3  c  ln
3. ln x  5  ln x  5  c  ln x 2  25  c
6. ln x  1  ln x  1  ln x  c  ln
7. ln x  1  ln x  1 
4. 1 ln 5
2
2
( x  1) 5 ( x  2) 2
( x  3) 7
c
5. 2 ln 2 

4
x2 1
c
x
2
x 1
2
 c  ln

c
x 1
x 1 x 1
 5  35 
  4  ln 1215   4
8. ln 5  5 ln 3  7 ln 2  4  ln
 27 
 128 


12
x2
12
9. 7 ln x  2  7 ln x  3 
 c  7 ln

c
x3
x3 x3
10.
3( 2  ln 2)
4
These exercises were taken from Mathematics Worksheets originally created by Study Advice Service
at Hull University. Web: www.hull.ac.uk/studyadvice Email: studyadvice@hull.ac.uk
Many thanks for use of these materials. Any comments can be sent to the email above or to
mathsskills@strath.ac.uk
7