7.4 Integration of Rational Functions by Partial Fractions TECHNIQUES OF INTEGRATION In this section, we will learn: How to integrate rational functions by reducing them to a sum of simpler fractions.
7.4
Integration of Rational Functions
by Partial Fractions
TECHNIQUES OF INTEGRATION
In this section, we will learn:
How to integrate rational functions
by reducing them to a sum of simpler fractions.
PARTIAL FRACTIONS
We show how to integrate any rational function by
expressing it as a sum of simpler fractions, called
partial fractions.
We already know how to integrate partial functions.
To illustrate the method, observe that, by taking
the fractions 2/(x – 1) and 1/(x – 2) to a common
denominator, we obtain:
INTEGRATION BY PARTIAL FRACTIONS
2
2 1 2( 2) ( 1)
1 2 ( 1)( 2)
5
2
x x
x x x x
x
x x
If we now reverse the procedure, we see how to
integrate the function on the right side of this
equation:
INTEGRATION BY PARTIAL FRACTIONS
2
5 2 1
2 1 2
2ln | 1| ln | 2 |
xdx dx
x x x x
x x C
To see how the method of partial fractions works
in general, let us consider a rational function
where P and Q are polynomials. It is 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.
INTEGRATION BY PARTIAL FRACTIONS
( )( )
( )
P xf x
Q x
Recall that, if
where an 0, then the degree of P is n and we write
deg(P) = n.
DEGREE OF P
1
11 0( ) n n
n nP x a x a x a x a
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).
PARTIAL FRACTIONS
The division statement is
where S and R are also polynomials.
As the following example illustrates, sometimes,
this preliminary step is all that is required.
PARTIAL FRACTIONS
( ) ( )( ) ( )
( ) ( )
P x R xf x S x
Q x Q x
Equation 1
Find
The degree of the numerator is greater than that of the denominator.
So, we first perform the long division.
PARTIAL FRACTIONS Example 1
3
1
x xdx
x
PARTIAL FRACTIONS
This enables us to write:
32
3 2
22
1 1
2 2ln | 1|3 2
x xdx x x dx
x x
x xx x C
Example 1
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.
FACTORISATION OF Q(x)
For instance, if Q(x) = x4 – 16, we could factor it
as:2 2
2
( ) ( 4)( 4)
( 2)( 2)( 4)
Q x x x
x x x
The third step is to express the proper rational
function R(x)/Q(x) as a sum of partial fractions
of the form:
FACTORISATION OF Q(x)
2or
( ) ( )i j
A Ax B
ax b ax bx c
A theorem in algebra guarantees that it is always
possible to do this.
We explain the details for the four cases that occur.
FACTORISATION OF Q(x)
The denominator Q(x) is a product of distinct
linear factors.
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.
CASE 1
In this case, the partial fraction theorem states that
there exist constants A1, A2, . . . , Ak such that:
These constants can be determined as in the
following example.
CASE 1
1 2
1 1 2 2
( )
( )k
k k
AA AR x
Q x a x b a x b a x b
Equation 2
Evaluate
The degree of the numerator is less than the degree of the denominator.
So, we do not need to divide.
PARTIAL FRACTIONS Example 2
2
3 2
2 1
2 3 2
x xdx
x x x
PARTIAL FRACTIONS
We factor the denominator as:
2x3 + 3x2 – 2x = x(2x2 + 3x – 2)
= x(2x – 1)(x + 2)
It has three distinct linear factors.
Example 2
Therefore, the partial fraction decomposition of the
integrand (Equation 2) has the form
PARTIAL FRACTIONS
2 2 1
(2 1)( 2) 2 1 2
x x A B C
x x x x x x
E. g. 2—Equation 3
To determine the values of A, B, and C, we multiply
both sides of the equation by the product of the
denominators, that is, by x(2x – 1)(x + 2), obtaining:
x2 + 2x + 1 = A(2x – 1)(x + 2) + Bx(x + 2)
+ Cx(2x – 1)
PARTIAL FRACTIONS E. g. 2—Equation 4
Expanding the right hand 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
PARTIAL FRACTIONS E. g. 2—Equation 5
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.
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
Solving, we get:
A = ½ B = 1/5 C = –1/10
PARTIAL FRACTIONS Example 2
Hence,
PARTIAL FRACTIONS
2
3 2
1 1 12 10 10
2 1
2 3 21 1 1 1 1 1
2 5 2 1 10 2
ln | | ln | 2 1| ln | 2 |
x xdx
x x x
dxx x x
x x x K
Example 2
PARTIAL FRACTIONS
In integrating the middle term, we have made the
mental substitution
u = 2x – 1, which gives
du = 2 dx and dx = du/2.
Example 2
We can use an alternative method to find the
coefficients A, B, and C in Example 2.
Equation 4 is an identity. It is true for every value
of x.
Let us choose values of x that simplify the equation.
NOTE
NOTE
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.
Likewise, x = ½ gives 5B/4 = 1/4 and x = –2
gives 10C = –1. Hence, B = 1/5 and C = –1/10.
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?
In fact, Equation 4 is true for all values of x, even
x = 0, ½, and –2 .
NOTE
Find , where a ≠ 0.
The method of partial fractions gives:
Therefore,
PARTIAL FRACTIONS Example 3
2 2
dx
x a
2 2
1 1
( )( )
A B
x a x a x a x a x a
( ) ( ) 1A x a B x a
We use the method of the preceding note.
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).
PARTIAL FRACTIONS Example 3
PARTIAL FRACTIONS
Therefore,
2 2
1 1 1
2
1(ln | | ln | |)
21
ln2
dxdx
x a a x a x a
x a x a Ca
x aC
a x a
Example 3
Q(x) is a product of linear factors, some of which
are repeated.
Suppose the first linear factor (a1x + b1) is repeated
r times.
That is, (a1x + b1)r occurs in the factorization of Q(x).
CASE 2
Then, instead of the single term A1/(a1x + b1)
in Equation 2, we would use:
CASE 2
1 22
1 1 1 1 1 1 ( ) ( )r
r
A A A
a x b a x b a x b
Equation 7
By way of illustration, we could write:
However, we prefer to work out in detail a simpler example, as follows.
CASE 2
3
2 3 2 2 3
1
( 1) 1 ( 1) ( 1)
x x A B C D E
x x x x x x x
Find
The first step is to divide.
The result of long division is:
PARTIAL FRACTIONS Example 4
4 2
3 2
2 4 1
1
x x xdx
x x x
4 2
3 2 3 2
2 4 1 41
1 1
x x x xx
x x x x x x
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:
PARTIAL FRACTIONS
3 2 2
2
1 ( 1)( 1)
( 1)( 1)( 1)
( 1) ( 1)
x x x x x
x x x
x x
Example 4
The linear factor x – 1 occurs twice. Therefore, the
partial fraction decomposition is:
PARTIAL FRACTIONS
2 2
4
( 1) ( 1) 1 ( 1) 1
x A B C
x x x x x
Example 4
Multiplying by the least common denominator,
(x – 1)2 (x + 1), we get:
PARTIAL FRACTIONS
2
2
4 ( 1)( 1) ( 1) ( 1)
( ) ( 2 ) ( )
x A x x B x C x
A C x B C x A B C
E. g. 4—Equation 8
PARTIAL FRACTIONS
If we equate coefficients, we get the linear system:
Solving, we obtain:
A = 1 B = 2 C = -1
0
2 4
0
A C
B C
A B C
Example 4
PARTIAL FRACTIONS
Thus, 4 2
3 2
2
2
2
2 4 1
1
1 2 11
1 ( 1) 1
2ln | 1| ln | 1|
2 1
2 1ln
2 1 1
x x xdx
x x x
x dxx x x
xx x x K
x
x xx K
x x
Example 4
Q(x) contains irreducible quadratic factors, none of
which is repeated.
That is, Q(x) has a factor of the form ax2 + bx + c,
where b2 – 4ac < 0.
CASE 3
Then, in addition to the partial fractions given by
Equations 2 and 7, the expression for R(x)/Q(x)
will have a term of the form
where A and B are constants to be determined.
CASE 3 Formula 9
2
Ax B
ax bx c
For instance, the function given by
f (x) = x/[(x – 2)(x2 + 1)(x2 + 4)]
has a partial fraction decomposition of the form
CASE 3
2 2
2 2
( 2)( 1)( 4)
2 1 4
x
x x x
A Bx C Dx E
x x x
The term in Formula 9 can be integrated by
completing the square and using the formula
CASE 3
12 2
1tan
du uC
u a a a
Formula 10
Evaluate
As x3 + 4x = x(x2 + 4) can not be factored further, we write:
PARTIAL FRACTIONS Example 5
2
3
2 4
4
x xdx
x x
2
2 2
2 4
( 4) 4
x x A Bx C
x x x x
Multiplying by x(x2 + 4), we have:
PARTIAL FRACTIONS
2 2
2
2 4 ( 4) ( )
( ) 4
x x A x Bx C x
A B x Cx A
Example 5
PARTIAL FRACTIONS
Equating coefficients, we obtain:
A + B = 2 C = –1 4A = 4
Thus, A = 1, B = 1, and C = –1.
Example 5
In order to integrate the second term, we split it
into two parts:
We make the substitution u = x2 + 4 in the first of
these integrals so that du = 2x dx.
PARTIAL FRACTIONS
2 2 2
1 1
4 4 4
x xdx dx dx
x x x
Example 5
We evaluate the second integral by means of
Formula 10 with a = 2:
PARTIAL FRACTIONS
2
2
2 2
2 11 12 2
2 4
( 4)
1 1
4 4
ln | | ln( 4) tan ( / 2)
x xdx
x x
xdx dx dx
x x x
x x x K
Example 5
Evaluate
The degree of the numerator is not less than the degree of the denominator.
So, we first divide and obtain:
PARTIAL FRACTIONS
2
2
4 3 2
4 4 3
x xdx
x x
2
2 2
4 3 2 11
4 4 3 4 4 3
x x x
x x x x
Example 6
Notice that the quadratic 4x2 – 4x + 3 is irreducible
because its discriminant is b2 – 4ac = –32 < 0.
This means it can not be factored.
So, we do not need to use the partial fraction technique.
PARTIAL FRACTIONS Example 6
To integrate the function, we complete the square
in the denominator:
This suggests we make the substitution u = 2x – 1.
Then, du = 2 dx, and x = ½(u + 1).
PARTIAL FRACTIONS
2 24 4 3 (2 1) 2x x x
Example 6
Thus,
PARTIAL FRACTIONS
2
2 2
121
2 2
14 2
4 3 2 11
4 4 3 4 4 3
( 1) 1
21
2
x x xdx dx
x x x x
ux du
uu
x duu
Example 6
PARTIAL FRACTIONS
1 14 42 2
2 118
2 118
1
2 2
1 1ln( 2) tan
4 2 2
1 2 1ln(4 4 3) tan
4 2 2
ux du du
u u
ux u C
xx x x C
Example 6
Example 6 illustrates the general procedure for
integrating a partial fraction of the form
NOTE
22
where 4 0Ax B
b acax bx c
We complete the square in the denominator and
then make a substitution that brings the integral
into the form
Then, the first integral is a logarithm and the second is expressed in terms of tan-1.
NOTE
2 2 2 2 2 2
1Cu D udu C du D du
u a u a u a
Q(x) contains a repeated irreducible quadratic
factor.
Suppose Q(x) has the factor (ax2 + bx + c)r
where b2 – 4ac < 0.
CASE 4
Then, instead of the single partial fraction
(Formula 9), the sum
occurs in the partial fraction decomposition
of R(x)/Q(x).
CASE 4
1 1 2 22 2 2 2 ( ) ( )
r rr
A x B A x B A x B
ax bx c ax bx c ax bx c
Formula 11
Write out the form of the partial fraction
decomposition of the function
PARTIAL FRACTIONS Example 7
3 2
2 2 3
1
( 1)( 1)( 1)
x x
x x x x x
We have:
PARTIAL FRACTIONS
3 2
2 2 3
2 2
2 2 2 3
1
( 1)( 1)( 1)
1 1 1
( 1) ( 1)
x x
x x x x x
A B Cx D Ex F
x x x x xGx h Ix J
x x
Example 7
Evaluate
The form of the partial fraction decomposition is:
PARTIAL FRACTIONS
2 3
2 2
1 2
( 1)
x x xdx
x x
2 3
2 2 2 2 2
1 2
( 1) 1 ( 1)
x x x A Bx C Dx E
x x x x x
Example 8
Multiplying by x(x2 + 1)2, we have:
PARTIAL FRACTIONS
3 2
2 2 2
4 2 4 2 3 2
4 3 2
2 1
( 1) ( ) ( 1) ( )
( 2 1) ( ) ( )
( ) (2 ) ( )
x x x
A x Bx C x x Dx E x
A x x B x x C x x Dx Ex
A B x Cx A B D x C E x A
Example 8
If we equate coefficients, we get the system
This has the solution A = 1, B = –1, C = –1, D = 1, E = 0.
PARTIAL FRACTIONS
0
1
2 2
1
1
A B
C
A B D
C E
A
Example 8
Thus,
PARTIAL FRACTIONS
2 3
2 2 2 2 2
2 2 2 2
2 112 2 2
1 2 1 1
( 1) 1 ( 1)
1 1 ( 1)
1ln | | ln( 1) tan
2( 1)
x x x x xdx dx
x x x x x
dx x dx x dxdx
x x x x
x x x Kx
Example 8
We note that, sometimes, partial fractions can be
avoided when integrating a rational function.
AVOIDING PARTIAL FRACTIONS
For instance, the integral
could be evaluated by the method of Case 3.
AVOIDING PARTIAL FRACTIONS
2
2
1
( 3)
xdx
x x
However, it is much easier to observe that,
if u = x(x2 + 3) = x3 + 3x, then du = (3x2 + 3) dx
and so
AVOIDING PARTIAL FRACTIONS
231
32
1ln | 3 |
( 3)
xdx x x C
x x
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 , then the substitution may be effective.
RATIONALIZING SUBSTITUTIONS
( )n g x ( )nu g x
Evaluate
Let
Then, u2 = x + 4
So, x = u2 – 4 and dx = 2u du
RATIONALIZING SUBSTITUTIONS Example 9
4xdx
x
4u x
We can evaluate this integral by factoring u2 – 4 as
(u – 2)(u + 2) and using partial fractions.
RATIONALIZING SUBSTITUTIONS Example 9