# Integration of Rational Functions by Partial Fractions Contributed by: In this section, we will learn: How to integrate rational functions by reducing them to a sum of simpler fractions.
1. 7
TECHNIQUES OF INTEGRATION
2. TECHNIQUES OF INTEGRATION
7.4
Integration of Rational Functions
by Partial Fractions
In this section, we will learn:
How to integrate rational functions
by reducing them to a sum of simpler fractions.
3. PARTIAL FRACTIONS
We show how to integrate any rational
function (a ratio of polynomials) by
expressing it as a sum of simpler fractions,
called partial fractions.
 We already know how to integrate
partial functions.
4. INTEGRATION BY PARTIAL FRACTIONS
To illustrate the method, observe that,
by taking the fractions 2/(x – 1) and 1/(x – 2)
to a common denominator, we obtain:
2 1 2( x  2)  ( x  1)
 
x 1 x2 ( x  1)( x  2)
x 5
 2
x x 2
5. INTEGRATION BY PARTIAL FRACTIONS
If we now reverse the procedure, we see
how to integrate the function on the right side
of this equation:
x 5  2 1 
x 2  x  2 dx  x  1  x  2  dx
2 ln | x  1|  ln | x  2 |  C
6. INTEGRATION BY PARTIAL FRACTIONS
To see how the method of partial fractions
works in general, let’s consider a rational
P( x)
f ( x) 
Q( x)
where P and Q are polynomials.
7. PROPER FUNCTION
It’s possible to express f as a sum of
simpler fractions if the degree of P is less
than the degree of Q.
Such a rational function is called proper.
8. DEGREE OF P
Recall that, if
n n 1
P( x) an x  an  1 x   a1 x  a0
where an ≠ 0, then the degree of P is n
and we write deg(P) = n.
9. PARTIAL FRACTIONS
If f is improper, that is, deg(P) ≥ deg(Q), then
we must take the preliminary step of dividing
Q into P (by long division).
 This is done until a remainder R(x) is obtained
such that deg(R) < deg(Q).
10. PARTIAL FRACTIONS Equation 1
The division statement is
P ( x) R( x)
f ( x)  S ( x ) 
Q( x) Q( x)
where S and R are also polynomials.
11. PARTIAL FRACTIONS
As the following example illustrates,
sometimes, this preliminary step is all
that is required.
12. PARTIAL FRACTIONS Example 1
3
x x
Find  dx
x 1
 The degree of the numerator is greater
than that of the denominator.
 So, we first perform the long division.
13. PARTIAL FRACTIONS Example 1
 This enables us to write:
3
x x  2 2 
x  1 dx 
 x  x  2   dx
x  1
3 2
x x
   2 x  2 ln | x  1|  C
3 2
14. PARTIAL FRACTIONS
The next step is to factor
the denominator Q(x) as far
as possible.
15. FACTORISATION OF Q(x)
It can be shown that any polynomial Q
can be factored as a product of:
 Linear factors (of the form ax + b)
 Irreducible quadratic factors (of the form ax2 + bx + c,
where b2 – 4ac < 0).
16. FACTORISATION OF Q(x)
For instance, if Q(x) = x4 – 16, we could
factor it as:
2 2
Q ( x) ( x  4)( x  4)
2
( x  2)( x  2)( x  4)
17. FACTORISATION OF Q(x)
The third step is to express the proper rational
function R(x)/Q(x) as a sum of partial fractions
of the form:
A Ax  B
i
or 2 j
(ax  b) (ax  bx  c)
18. FACTORISATION OF Q(x)
A theorem in algebra guarantees that
it is always possible to do this.
 We explain the details for the four cases
that occur.
19. CASE 1
The denominator Q(x)
is a product of distinct linear
20. CASE 1
This means that we can write
Q(x) = (a1x + b1) (a2x + b2)…(akx + bk)
where no factor is repeated (and no factor
is a constant multiple of another.
21. CASE 1 Equation 2
In this case, the partial fraction theorem states
that there exist constants A1, A2, . . . , Ak such
R( x) A1 A2 Ak
   
Q( x) a1 x  b1 a2 x  b2 ak x  bk
22. CASE 1
These constants can be
determined as in the following
23. PARTIAL FRACTIONS Example 2
2
x  2x  1
Evaluate  3 2
dx
2 x  3x  2 x
 The degree of the numerator is less than
the degree of the denominator.
 So, we don’t need to divide.
24. PARTIAL FRACTIONS Example 2
We factor the denominator as:
2x3 + 3x2 – 2x = x(2x2 + 3x – 2)
= x(2x – 1)(x + 2)
 It has three distinct linear factors.
25. PARTIAL FRACTIONS E. g. 2—Equation 3
So, the partial fraction decomposition of
the integrand (Equation 2) has the form
2
x  2x  1 A B C
  
x(2 x  1)( x  2) x 2 x  1 x  2
26. PARTIAL FRACTIONS E. g. 2—Equation 4
To determine the values of A, B, and C, we
multiply both sides of the equation by the
product of the denominators, x(2x – 1)(x + 2),
x2 + 2x + 1 = A(2x – 1)(x + 2) + Bx(x + 2)
+ Cx(2x – 1)
27. PARTIAL FRACTIONS E. g. 2—Equation 5
Expanding the right side of Equation 4 and
writing it in the standard form for polynomials,
we get:
x2 + 2x + 1 = (2A + B + 2C)x2
+ (3A + 2B – C) – 2A
28. PARTIAL FRACTIONS Example 2
The polynomials in Equation 5 are identical.
So, their coefficients must be equal.
 The coefficient of x2 on the right side, 2A + B + 2C,
must equal that of x2 on the left side—namely, 1.
 Likewise, the coefficients of x are equal and
the constant terms are equal.
29. PARTIAL FRACTIONS Example 2
This gives the following system of equations
for A, B, and C:
2A + B + 2C = 1
3A + 2B – C = 2
–2A = –1
30. PARTIAL FRACTIONS Example 2
Solving, we get:
A=½
 B = 1/5
 C = –1/10
31. PARTIAL FRACTIONS Example 2
2
x  2x  1
2 x3  3x 2  2 x dx
11 1 1 1 1 
    dx
 2 x 5 2 x  1 10 x  2 
 12 ln | x |  101 ln | 2 x  1|  101 | x  2 |  K
32. PARTIAL FRACTIONS Example 2
In integrating the middle term,
we have made the mental substitution
u = 2x – 1, which gives
du = 2 dx and dx = du/2.
33. We can use an alternative method
to find the coefficients A, B, and C
in Example 2.
34. Equation 4 is an identity.
It is true for every value of x.
 Let’s choose values of x that simplify
the equation.
35. If we put x = 0 in Equation 4, the second
and third terms on the right side vanish, and
the equation becomes –2A = –1.
 Hence, A = ½.
36. Likewise, x = ½ gives 5B/4 = 1/4
and x = –2 gives 10C = –1.
 Hence, B = 1/5 and C = –1/10.
37. You may object that Equation 3 is not
valid for x = 0, ½, or –2.
 So, why should Equation 4 be valid for
those values?
38. In fact, Equation 4 is true for all values
of x, even x = 0, ½, and –2 .
 See Exercise 69 for the reason.
39. PARTIAL FRACTIONS Example 3
dx
Find  2 2
, where a ≠ 0.
x a
 The method of partial fractions gives:
1 1 A B
2 2
  
x  a ( x  a)( x  a ) x  a x  a
 Therefore, A( x  a)  B ( x  a) 1
40. PARTIAL FRACTIONS Example 3
We use the method of the preceding
 We put x = a in the equation and get A(2a) = 1.
So, A = 1/(2a).
 If we put x = –a, we get B(–2a) = 1.
So, B = –1/(2a).
41. PARTIAL FRACTIONS Example 3
dx 1  1 1 
2 2
  
x  a 2a  x  a x  a   dx
1
 (ln | x  a |  ln | x  a |)  C
2a
42. PARTIAL FRACTIONS E. g. 3—Formula 6
Since ln x – ln y = ln(x/y), we can write
the integral as:
dx 1 x a
x 2  a 2  2a ln x  a  C
 See Exercises 55–56 for ways of using Formula 6.
43. CASE 2
Q(x) is a product of
linear factors, some of which
are repeated.
44. CASE 2
Suppose the first linear factor (a1x + b1) is
repeated r times.
 That is, (a1x + b1)r occurs in the factorization
of Q(x).
45. CASE 2 Equation 7
Then, instead of the single term A1/(a1x + b1)
in Equation 2, we would use:
A1 A2 Ar
 2
 r
a1 x  b1 (a1 x  b1 ) (a1 x  b1 )
46. CASE 2
By way of illustration, we could write:
3
x  x 1 A B C D E
2 3
  2  2
 3
x ( x  1) x x x  1 ( x  1) ( x  1)
 However, we prefer to work out in detail
a simpler example, as follows.
47. PARTIAL FRACTIONS Example 4
4 2
x  2 x  4 x 1
Find  3 2 dx
x  x  x 1
 The first step is to divide.
4 2
 The result of long division is:
x  2 x  4 x 1
x3  x 2  x  1
4x
x  1  3
x  x2  x 1
48. PARTIAL FRACTIONS Example 4
The second step is to factor the
denominator Q(x) = x3 – x2 – x + 1.
 Since Q(1) = 0, we know that x – 1 is a factor,
and we obtain:
x3  x 2  x  1 ( x  1)( x 2  1)
( x  1)( x  1)( x  1)
2
( x  1) ( x  1)
49. PARTIAL FRACTIONS Example 4
The linear factor x – 1 occurs twice.
So, the partial fraction decomposition is:
4x A B C
2
  2

( x  1) ( x  1) x  1 ( x  1) x 1
50. PARTIAL FRACTIONS E. g. 4—Equation 8
Multiplying by the least common denominator,
(x – 1)2 (x + 1), we get:
2
4 x  A( x  1)( x  1)  B( x  1)  C ( x  1)
2
( A  C ) x  ( B  2C ) x  ( A  B  C )
51. PARTIAL FRACTIONS Example 4
Now, we equate coefficients:
A  C 0
B  2C 4
 A  B  C 0
52. PARTIAL FRACTIONS Example 4
Solving, we obtain:
A=1
B=2
C = -1
53. PARTIAL FRACTIONS Example 4
Thus, 4 2
x  2 x  4 x 1
 x3  x 2  x 1 dx
 1 2 1 
 x  1   2
  dx
 x  1 ( x  1) x 1
x2 2
  x  ln | x  1|   ln | x  1|  K
2 x 1
2
x 2 x 1
 x  ln K
2 x 1 x 1
54. CASE 3
Q(x) contains irreducible
is repeated.
55. CASE 3 Formula 9
If Q(x) has the factor ax2 + bx + c, where
b2 – 4ac < 0, then, in addition to the partial
fractions in Equations 2 and 7, the expression
for R(x)/Q(x) will have a term of the form
Ax  B
2
ax  bx  c
where A and B are constants to be
56. CASE 3
For instance, the function given by
f(x) = x/[(x – 2)(x2 + 1)(x2 + 4) has a partial
fraction decomposition of the form
x
2 2
( x  2)( x  1)( x  4)
A Bx  C Dx  E
  2  2
x  2 x 1 x 4
57. CASE 3 Formula 10
The term in Formula 9 can be integrated
by completing the square and using
the formula
dx 1 1  x 
x 2  a 2  a tan  a   C
58. PARTIAL FRACTIONS Example 5
2
2
Evaluate  3x  x  4
dx
x  4x
 As x3 + 4x = x(x2 + 4) can’t be factored further,
we write: 2
2 x  x  4 A Bx  C
2
  2
x( x  4) x x 4
59. PARTIAL FRACTIONS Example 5
Multiplying by x(x2 + 4), we have:
2 2
2 x  x  4  A( x  4)  ( Bx  C ) x
2
( A  B) x  Cx  4 A
60. PARTIAL FRACTIONS Example 5
Equating coefficients, we obtain:
A+B=2 C = –1 4A = 4
 Thus, A = 1, B = 1, and C = –1.
61. PARTIAL FRACTIONS Example 5
2
2x  x  4  1 x 1 
 x3  4 x dx  
 x x 2  4 
  dx
62. PARTIAL FRACTIONS Example 5
In order to integrate the second term,
we split it into two parts:
x 1 x 1
x 2  4 dx x 2  4 dx  x2  4 dx
63. PARTIAL FRACTIONS Example 5
We make the substitution u = x2 + 4
in the first of these integrals so that
du = 2x dx.
64. PARTIAL FRACTIONS Example 5
We evaluate the second integral by means
of Formula 10 with a = 2:
2
2x  x  4
 x( x 2  4) dx
1 x 1
 dx   2 dx   2 dx
x x 4 x 4
2 1
ln | x |  2 ln( x  4)  2 tan ( x / 2)  K
1 1
65. PARTIAL FRACTIONS Example 6
2
4 x  3x  2
Evaluate  2 dx
4x  4x  3
 The degree of the numerator is not less than
the degree of the denominator.
4 x 2  3x  2
 So, we first divide and obtain:
4x2  4x  3
x 1
1  2
4x  4x  3
66. PARTIAL FRACTIONS Example 6
Notice that the quadratic 4x2 – 4x + 3
is irreducible because its discriminant
is b2 – 4ac = –32 < 0.
 This means it can’t be factored.
 So, we don’t need to use the partial fraction technique.
67. PARTIAL FRACTIONS Example 6
To integrate the function, we complete
the square in the denominator:
2 2
4 x  4 x  3 (2 x  1)  2
 This suggests we make the substitution u = 2x – 1.
 Then, du = 2 dx, and x = ½(u + 1).
68. PARTIAL FRACTIONS Example 6
2
4 x  3x  2
4 x 2  4 x  3 dx
 x 1 
 1  2  dx
 4x  4x  3 
2 (u  1)  1
1
x  2  2
1
du
u 2
u 1
x  4  2
1
du
u 2
69. PARTIAL FRACTIONS Example 6
u 1
x   2
1
4 du  4  2
1
du
u 2 u 2
2 1 1 1  u 
 x  8 ln(u  2)   tan 
1
 C
4 2  2
2 1  1  2x  1 
 x  8 ln(4 x  4 x  3) 
1
tan   C
4 2  2 
70. Example 6 illustrates the general
procedure for integrating a partial fraction
of the form
Ax  B 2
2
where b  4ac  0
ax  bx  c
71. We complete the square in the denominator
and then make a substitution that brings
the integral into the form
Cu  D u 1
u 2  a 2 du C u 2  a 2 du  D u 2  a 2 du
 Then, the first integral is a logarithm and
the second is expressed in terms of tan-1.
72. CASE 4
Q(x) contains
a repeated irreducible
73. CASE 4
Suppose Q(x) has the factor
(ax2 + bx + c)r
where b2 – 4ac < 0.
74. CASE 4 Formula 11
Then, instead of the single partial fraction
(Formula 9), the sum
A1 x  B1 A2 x  B2 Ar x  Br
2
 2 2

ax  bx  c (ax  bx  c) (ax 2  bx  c )r
occurs in the partial fraction decomposition
of R(x)/Q(x).
75. CASE 4
Each of the terms in Formula 11
can be integrated by first completing
the square.
76. PARTIAL FRACTIONS Example 7
Write out the form of the partial fraction
decomposition of the function
3 2
x  x 1
2 2 3
x( x  1)( x  x  1)( x  1)
77. PARTIAL FRACTIONS Example 7
We have:
3 2
x  x 1
2 2 3
x( x  1)( x  x  1)( x  1)
A B Cx  D Ex  F
   2  2
x x  1 x  x 1 x 1
Gx  h Ix  J
 2 2
 2 3
( x  1) ( x  1)
78. PARTIAL FRACTIONS Example 8
2 3
1 x  2x  x
Evaluate  2 2
dx
x( x  1)
 The form of the partial fraction decomposition is:
2 3
1 x  2x  x A Bx  C Dx  E
2 2
  2  2 2
x( x  1) x x  1 ( x  1)
79. PARTIAL FRACTIONS Example 8
Multiplying by x(x2 + 1)2,
we have:
 x3  2 x 2  x  1
2 2 2
 A( x  1)  ( Bx  C ) x( x  1)  ( Dx  E ) x
 A( x 4  2 x 2  1)  B ( x 4  x 2 )  C ( x 3  x)  Dx 2  Ex
( A  B ) x 4  Cx 3  (2 A  B  D) x 2  (C  E ) x  A
80. PARTIAL FRACTIONS Example 8
If we equate coefficients,
we get the system A  B 0
C  1
2 A  B  D 2
C  E  1
A 1
 This has the solution
A = 1, B = –1, C = –1, D = 1, E =
0.
81. PARTIAL FRACTIONS Example 8
2 3
1 x  2x  x
 x( x 2 1)2 dx
 1 x 1 x 
  2  2 2 
dx
 x x  1 ( x  1) 
dx x dx x dx
   2 dx   2  2
x x 1 x 1 ( x  1) 2
2 1 1
ln | x |  2 ln( x  1)  tan x 
1
2 2
K
2( x  1)
82. AVOIDING PARTIAL FRACTIONS
We note that, sometimes,
partial fractions can be avoided
when integrating a rational function.
83. AVOIDING PARTIAL FRACTIONS
For instance, the integral
2
x 1
x( x 2  3) dx
could be evaluated by the method
of Case 3.
84. AVOIDING PARTIAL FRACTIONS
However, it is much easier to observe that,
if u = x(x2 + 3) = x3 + 3x, then du = (3x2 + 3) dx
and so
2
x 1 3
x( x 2  3) dx  3 ln | x  3x |  C
1
85. RATIONALIZING SUBSTITUTIONS
Some nonrational functions can be
changed into rational functions by means
of appropriate substitutions.
 In particular, when an integrand contains
an expression of the form n√g(x), then
the substitution u = n√g(x) may be effective.
86. RATIONALIZING SUBSTITUTIONS Example 9
Evaluate x4
 dx
x
 Let u  x4
 Then, u2 = x + 4
 So, x = u2 – 4 and dx = 2u du
87. RATIONALIZING SUBSTITUTIONS Example 9
 Therefore,
x4 u
 dx  2 2u du
x u 4
2
u
2 2 du
u 4
 4 
2  1  2  4  du
 u 
88. RATIONALIZING SUBSTITUTIONS Example 9
We can evaluate this integral
by factoring u2 – 4 as (u – 2)(u + 2)
and using partial fractions.
89. RATIONALIZING SUBSTITUTIONS Example 9
Alternatively, we can use Formula 6
with a = 2: x4
 x dx
du
2du  8 2
u 4
1 u 2
2u  8  ln C
2 2 u  2
x4  2
2 x  4  2 ln C
x4 2