techniques of integration

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