7.4 Integration of Rational Functions by Partial Fractions TECHNIQUES OF INTEGRATION In this...

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:


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.


( )( )

( )

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

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.


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:


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.


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.


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.


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


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,


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

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:


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:


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

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:

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.


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

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


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.


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


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

Therefore,


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:


2

42 8

41 2

2 8 ln2 2 2

4 22 4 2ln

4 2

x dudx du

x uu

u Cu

xx C

x

Example 9

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