Partial Fractions and Integration
Mathematics Skills Guide
This is one of a series of guides designed to help you increase your confidence in
handling mathematics. This guide contains both theory and exercises which cover:1. Partial fractions
2. Integration using partial fractions
There are often different ways of doing things in mathematics and the methods
suggested in the guides may not be the ones you were taught. If you are successful
and happy with the methods you use it may not be necessary for you to change
them. If you have problems or need help in any part of the work then there are a
number of ways you can get help.
For students at the University of Hull
 Ask your lecturers.
 You can contact a maths Skills Adviser from the Skills Team on the email shown
below.
 Access more maths Skills Guides and resources at the website below.
 Look at one of the many textbooks in the library.
Web: www.hull.ac.uk/skills
Email: skills@hull.ac.uk
1. Introduction (with some brief theory)
5
To evaluate 
dx we need to be able to split the fraction into what are
x  2x  3
called Partial Fractions.
Adding fractions
1
1


x  2 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
Let
5
x  2x  3
5
in partial fractions
x  2x  3

A
B

x2 x3
Multiplying through by ( x  2)( x  3) , gives 5  A( x  3)  B( x  2)
This is an identity, true for every value of x .
This means that we can take any value of x and substitute it into both sides of the
identity to find the values of A and B. The most convenient values are x  2 and
x  3 as each makes one bracket zero giving the value of one unknown in each
case.
Putting x  2 gives 5  A(2  3)  B(2  2)  A  1
Putting x  3 gives 5  A(3  3)  B(3  2)  B  1
5
1
1
So


x  2x  3 x  2 x  3
Which is where we started!
Notice that, as we are dealing with an identity, when we work out the coefficient of,
say, x it must be the same on both sides.
From 5  Ax  3  Bx  2
we get 5  A  Bx  3A  2B
equate coefficient of x 0  A  B
(i)
equate the constant terms 5  3A  2B
(ii)
add equn (ii) to 2 × equn (i) 5  5 A  A  1
substituting into (i) or (ii) gives
B  1
1
Note that we get A  1 , B  1 as before and as expected!
In this case the first method is much easier to use but the second method will have to
be used when dealing with more complex functions. Adding the 2 fractions in this
section helps to decide what we expect to happen when we reverse the process.
The following exercise will help sort some more ideas out.
Exercise 1
Simplify the following ie express them as single fractions
2x
3
1
2
2.

1.

2
2x  1
 x  1
 2 x  1
x 1
3.
1

x 1
1
 x  12

1
 x  13
4.
3

 x  1
5
2
 x  3
2. Some Partial Fractions
(i) Proper fractions (ie fractions where the highest power in the numerator is less
than the highest power in the denominator)
x 2  9 x  25
x  12 x  1x  5
is a proper fraction as the highest power in the numerator is 2 while, if the brackets
were multiplied out, the highest power in the denominator would be 3, coming from
x  2 x  x  2 x3 .
x3
is improper as the highest power in both numerator
x  12 x  1x  5
and denominator is 3.
The fraction
x 4  3x 2  x  1
is improper as the highest power in numerator (4) is
x  12 x  1x  5
greater than the highest power in the denominator (3).
The fraction
From dealing with addition and subtraction of algebraic fractions various deductions
can be made. In the following the highest power in the function f ( x ) is lower than
the highest power in the denominator i.e. 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



2
2
x  1 2 x  1x  5 x  1 2 x  1 x  5


2
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
.
Note that number 4 in the exercise leads to an improper fraction as the highest power
in both numerator and denominator is 2.
The ideas introduced here can go on to repeated quadratic factors, to repeated and
non-factorisable cubic factors etc. This worksheet only deals with the 3 types
mentioned above but the methods can be extended to deal with such expressions as
3 x3  3 x 2  2 x  7
Ax  B
Cx  D
x2  2 x  3 x2  2 x
2
(a) - Type 1 Express
Let

2

2
2


E
x  3
3x  1
in partial fractions
x  12 x  1
3x  1
A
B


x  12 x  1 x  1 2 x  1
Multiplying through by x  12 x  1
gives
put x  1
3x  1  A2 x  1  Bx  1
3  1  A2  1  0
1 A B
put x  0 (equate constant terms)
 A 4
B
3
1
3
RHS = 2 A + B  8  1  3  LHS)
(check coeff of x
3
3
Note it would have been possible to substitute x   1 giving B directly:-
 2


2
3  1 1  B  1 1  B  1
2
3
but the above method avoids the fractions!
3x  1
4
1
Solution


x  12 x  1 3x  1 32 x  1
3
(b) - Type 2 Express
3x  1
x  1x2  1
3x  1
Let
Multiplying through by
gives
putting x  1
x  1x2  1
(c) - Type 3 Express

2
 2x  1
2
2x  1



x  1 x2  1
x  1 x2  1

36
x  12 x  5
36
Let

0  A  B  B  2
1 AC  C 1
RHS =  B + C  (2)  1  3  LHS)
equate constant term
(check coeff of x
3x  1
A
Bx  C

x  1 x2  1
 
x  1x2  1
x  1x2  1
3x  1  Ax 2  1  Bx  C x  1
3  1  A1  1  A  2
equate coefficient of x 2
Solution
in partial fractions



in partial fractions
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 x  5 x  1
2
2

1
1

x  1 x  5
(ii) Improper Fractions
The fraction
x 4  2 x3  x 2  4 x  4
x  3x 2  1
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
 x 1
x2  2 x  7
x  3x 2  1
4
x2  2 x  7
x  3x2  1
The fraction
giving
can then be put into partial fractions as in example 3 above
x 4  2 x3  x 2  4 x  4
x  3x2  1
 x 1
Exercise 2
Express the following in Partial Fractions
3x
2x  1
1.
2.
x  2x  1
x  1x  2x  3
4
3x 2  2 x
4.
5
.
x x2  4
x  2 x 2  4
2
x 2  3x
7.
2
8
.
x  1 x  1
x2  4
x4  1
1
10.
11
.
x x  1 x  1
x3  2 x




1
2

x  3 x 2  1

3.
6.
9.
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. 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
4 2x 2  3
4
2x  3
2x
1
2x  1
3. Splitting up such expressions as
to get

x2  1
x2  1 x2  1
2x  1
2x
1
dx  

dx  ln x 2  1  tan 1 x  c
giving 
2
2
2
x 1
x 1 x 1




Examples
(a) Simplify

3x  1
dx
x  12 x  1
From example. (a) on page 3

3x  1
4
dx 
3  x  1
 x  1 2 x  1



4
3
1
  x  1 dx
4
ln x  1
3

1
dx
3  2 x  1

1
6

1
ln 2 x  1  c
6
2
  2 x  1
dx
5
(b) Si mplify
3x  1
 x  1x 2  1 dx
From example. (b) on page 4
3x  1
  x  1


x2  1
dx 

2
 x 1
2
 x 1


1  2x
x2  1
1
2
x 1
dx

2x
2
x 1
dx


 2 ln  x  1  tan1 x  ln x 2  1  c
(c) Find the value of
4
 2 x  12 x  5dx
36
From example (c) on page 4 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
The answer can be left in this form unless the question specifies, say, 2 decimal
places.
6
Note: You should be able to do
dx “at sight”! If you can’t, then use
x  12
substitution.

6
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 
39 x  9
Expressing
2 x2  x  1

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
Let
39 x  9  A  2 x  1  B  x  1
then
putting x  1 gives 48  3 A  A  16
putting x  0 gives 9  A  B  B  7
24 x3  x  3
This gives
Hence
 12 x 2  30 x  9 
16
7

 x  1  2 x  1
 x  1 2 x  1
5 24 x3  x  3
5
16
7
2
 4  x  1 2 x  1 dx   4 12 x  30 x  9   x  1   2 x  1 dx
5
  4 x3  15 x 2  9 x  16 ln x  1  7 ln 2 x  1 
2

4
 500  375  45  16 ln 4  7 ln11  256  240  36  16 ln 3  7 ln 9

2
 
 80  32 ln 2  7 ln11   20  16 ln 3  7 ln 3
2

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
This could be “worked out” using a calculator if it is really necessary but this is the
exact answer and it depends on the context as to how much more you do, if
anything!
(e) Evaluate
1
x
 x2  1


x2 x2  1
1 x
2
x
2
1 x
First express
Writing
1 x
2
 x  1
2

A
x
2

dx
in Partial Fractions
B Cx  D

x
x2  1
and equating terms gives A  1 , B  1 , C  1 , D  1
7
1 x
2
Hence
1
x
2
 x  1
2
dx 

2
1
1
2
1
x
2

x 2 
1 x 1

dx
x x2  1
1 1  2x 
1
 

dx

x 2  x2  1  x2  1


2
   x 1  ln x  1 ln x 2  1  tan1 x 

 1
2

 

  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
x 2  25
2
4.
dx
3x 2  2 x
5.
0
7.
 x  12 x  1 dx
9.
x  2x 2  4
dx
4
8.
5x  3
 x  2x  32 dx
Answers
Exercise 1
3
1.
 x  1 2 x  1
3.
6.
x2  x  1
 x  13
10.
1 xx2  4 dx
4
x2  1
 xx2  1 dx



8 2 x 2  3x
dx
2
6
x 4
2 x4  1
dx
1 3

x  2x
7 x2  2 x  3
 x2  1  2x  1
2  x2  4
2.
4.
2
 x  1 x  3
8
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
We would appreciate your comments on this worksheet, especially if
you’ve found any errors, so that we can improve it for future use. Please
contact the Maths Skills Adviser by email at skills@hull.ac.uk
The information in this leaflet can be made available in an alternative format on
request using the email above.
9