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7TECHNIQUES OF INTEGRATIONTECHNIQUES OF INTEGRATION
7.4Integration 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 (a ratio of polynomials) 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’s consider a rational
function
where P and Q are polynomials.
INTEGRATION BY PARTIAL FRACTIONS
( )( )
( )
P xf x
Q x
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.
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.
PARTIAL FRACTIONS
( ) ( )( ) ( )
( ) ( )
P x R xf x S x
Q x Q x
Equation 1
PARTIAL FRACTIONS
As the following example illustrates,
sometimes, this preliminary step is all
that is required.
Find
The degree of the numerator is greater than that of the denominator.
So, we first perform the long division.
PARTIAL FRACTIONS Example 13
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
The next step is to factor
the denominator Q(x) as far
as possible.
PARTIAL FRACTIONS
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.
CASE 1
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.
In this case, the partial fraction theorem states
that there exist constants A1, A2, . . . , Ak such
that:
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
CASE 1
These constants can be
determined as in the following
example.
Evaluate
The degree of the numerator is less than the degree of the denominator.
So, we don’t need to divide.
PARTIAL FRACTIONS Example 22
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
So, 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, 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 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
PARTIAL FRACTIONS Example 2
PARTIAL FRACTIONS
Solving, we get:
A = ½
B = 1/5
C = –1/10
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| | 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.
NOTE
Equation 4 is an identity.
It is true for every value of x.
Let’s 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.
NOTE
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?
NOTE
NOTE
In fact, Equation 4 is true for all values
of x, even x = 0, ½, and –2 .
See Exercise 69 for the reason.
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 | |)
2
dx
dxx a a x a x a
x a x a Ca
Example 3
Since ln x – ln y = ln(x/y), we can write
the integral as:
See Exercises 55–56 for ways of using Formula 6.
PARTIAL FRACTIONS
2 2
1ln
2
dx x aC
x a a x a
E. g. 3—Formula 6
Q(x) is a product of
linear factors, some of which
are repeated.
CASE 2
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).
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 44 2
3 2
2 4 1
1
x x xdx
x x x
4 2
3 2
3 2
2 4 1
14
11
x x x
x x xx
xx 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.
So, 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
Now, we equate coefficients:
0
2 4
0
A C
B C
A B C
Example 4
Solving, we obtain:
A = 1
B = 2
C = -1
PARTIAL FRACTIONS 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 Kx x
Example 4
Q(x) contains irreducible
quadratic factors, none of which
is repeated.
CASE 3
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
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
dx xC
x a a a
Formula 10
Evaluate
As x3 + 4x = x(x2 + 4) can’t be factored further, we write:
PARTIAL FRACTIONS Example 52
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
Hence,
PARTIAL FRACTIONS
2
3 2
2 4 1 1
4 4
x x xdx dx
x x x x
Example 5
In order to integrate the second term,
we split it into two parts:
PARTIAL FRACTIONS
2 2 2
1 1
4 4 4
x xdx dx dx
x x x
Example 5
We make the substitution u = x2 + 4
in the first of these integrals so that
du = 2x dx.
PARTIAL FRACTIONS 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 dxx 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 FRACTIONS2
2
4 3 2
4 4 3
x xdx
x x
2
2
2
4 3 2
4 4 31
14 4 3
x x
x xx
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’t be factored.
So, we don’t 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) 2 x x x
Example 6
Thus,
PARTIAL FRACTIONS
2
2
2
121
2 2
14 2
4 3 2
4 4 31
14 4 3
( 1) 1
21
2
x xdx
x xx
dxx 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
u
x du duu 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.
CASE 4
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 r
r
A x B A x B A x B
ax bx c ax bx c ax bx c
Formula 11
CASE 4
Each of the terms in Formula 11
can be integrated by first completing
the square.
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 FRACTIONS2 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 xdx
x x
x xdx
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 n√g(x), then the substitution u = n√g(x) may be effective.
RATIONALIZING SUBSTITUTIONS
Evaluate
Let
Then, u2 = x + 4
So, x = u2 – 4 and dx = 2u du
RATIONALIZING SUBSTITUTIONS Example 9
4xdx
x
4u x
Therefore,
RATIONALIZING SUBSTITUTIONS
2
2
2
2
42
4
244
2 1 4
x udx u du
x u
udu
u
duu
Example 9
We can evaluate this integral
by factoring u2 – 4 as (u – 2)(u + 2)
and using partial fractions.
RATIONALIZING SUBSTITUTIONS Example 9
Alternatively, we can use Formula 6
with a = 2:
RATIONALIZING SUBSTITUTIONS
2
4
2 84
1 22 8 ln
2 2 2
4 22 4 2ln
4 2
xdx
xdu
duu
uu C
u
xx C
x
Example 9