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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 x2 ( 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
quadratic factors, none of which
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
quadratic factor.
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 x4
dx
x
Let u x4
Then, u2 = x + 4
So, x = u2 – 4 and dx = 2u du
87.
RATIONALIZING SUBSTITUTIONS Example 9
Therefore,
x4 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: x4
x dx
du
2du 8 2
u 4
1 u 2
2u 8 ln C
2 2 u 2
x4 2
2 x 4 2 ln C
x4 2
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