a lecture on the partial fraction decomposition method

a lecture on the partial fraction decomposition method

1

Integrating Rational Functions

A

rational function

is a function of the form

y

=

p

⁢

(

x

)

q

⁢

(

x

)

where

p

⁢

(

x

)

and

q

⁢

(

x

)

are

polynomials

(with real

coefficients

). Here we are interested in how to solve
the

integral

∫

p

⁢

(

x

)

q

⁢

(

x

)

⁢

𝑑

x

We already know how to integrate some functions of this type. As
we know, the

chain rule

tells us that the

derivative

of

y

=

ln

⁡

(

g

⁢

(

x

)

)

is

y

′

=

g

′

⁢

(

x

)

g

⁢

(

x

)

, where

g

⁢

(

x

)

is any other
function. Therefore:

∫

g

′

⁢

(

x

)

g

⁢

(

x

)

⁢

𝑑

x

=

ln

⁡

|

g

⁢

(

x

)

|

+

C

.

(1)

Example 1.1

.

∫

2

⁢

x

+

1

x

2

+

x

+

1

⁢

𝑑

x

=

ln

⁡

|

x

2

+

x

+

1

|

+

C

.

Example 1.2

.

Equation

(

1

) may also be used to integrate any function of
the form

y

=

a

b

⁢

x

+

c

, for any constants

a

,

b

,

c

. Indeed:

∫

a

b

⁢

x

+

c

⁢

𝑑

x

=

a

b

⁢

∫

b

b

⁢

x

+

c

⁢

𝑑

x

=

a

b

⁢

ln

⁡

|

b

⁢

x

+

c

|

+

C

or alternatively use a
substitution

u

=

b

⁢

x

+

c

.

Example 1.3

.

Using a substitution we may also integrate any function of the
form

y

=

a

(

b

⁢

x

+

c

)

n

, namely

u

=

b

⁢

x

+

c

does the job. For
example, using

u

=

2

⁢

x

+

1

,

d

⁢

u

=

2

⁢

d

⁢

x

:

∫

3

(

2

⁢

x

+

1

)

4

⁢

𝑑

x

=

3

2

⁢

∫

1

u

4

⁢

𝑑

u

=

-

1

2

⁢

1

u

3

+

C

=

-

1

2

⁢

(

2

⁢

x

+

1

)

3

+

C

.

Example 1.4

.

The derivative of the

arc tangent

function,

y

=

arctan

⁡

x

, is

y

′

=

1

1

+

x

2

. Therefore:

∫

1

1

+

x

2

⁢

𝑑

x

=

arctan

⁡

x

+

C

.

Thus, for any constant

a

, using a substitution

x

=

a

⁢

u

one may
integrate

∫

1

a

2

+

x

2

⁢

𝑑

x

=

1

a

⁢

arctan

⁡

x

a

+

C

.

You may also use the arctangent function to integrate other
functions, by completing the square of the denominator. For
example,

x

2

+

2

⁢

x

+

2

=

1

+

(

x

+

1

)

2

, thus:

∫

1

x

2

+

2

⁢

x

+

2

⁢

𝑑

x

=

∫

1

1

+

(

x

+

1

)

2

⁢

𝑑

x

=

arctan

⁡

(

x

+

1

)

+

C

.

Example 1.5

.

The arctangent also allows us to integrate another family of
functions, namely:

∫

b

⁢

x

+

c

a

2

+

x

2

⁢

𝑑

x

.

The trick is to use the favorite strategy of Napoleon:

divide
and conquer

, i.e. we break the fraction into a sum of two (here
we pick

a

=

1

for simplicity):

∫

b

⁢

x

+

c

1

+

x

2

⁢

𝑑

x

=

∫

b

⁢

x

1

+

x

2

⁢

𝑑

x

+

∫

c

1

+

x

2

⁢

𝑑

x

=

b

2

⁢

ln

⁡

|

1

+

x

2

|

+

c

⁢

arctan

⁡

x

+

C

.

2

Partial Fraction Decomposition

The objective of this method is to reduce any integral of the type

∫

p

⁢

(

x

)

q

⁢

(

x

)

⁢

𝑑

x

to a sum of integrals of the types
described in the previous

sections

. For example:

Example 2.1

.

We would like to solve the following integral:

∫

2

x

2

+

3

⁢

x

-

4

⁢

𝑑

x

.

First, we factor the denominator:

x

2

+

3

⁢

x

-

4

=

(

x

+

4

)

⁢

(

x

-

1

)

In order
to integrate, we are going to express the fraction in the
integrand as a sum:

2

x

2

+

3

⁢

x

-

4

=

A

x

+

4

+

B

x

-

1

for some constants

A

,

B

to be determined. The right hand side
(after realizing a common denominator) adds up to

A

⁢

(

x

-

1

)

+

B

⁢

(

x

+

4

)

(

x

+

4

)

⁢

(

x

-

1

)

. Therefore, in order to have an
equality we need the numerators to be equal:

A

⁢

(

x

-

1

)

+

B

⁢

(

x

+

4

)

=

2

(2)

for

all

values of

x

(i.e. this
should be an equality of polynomials). Thus, we can substitute
values of

x

and obtain equations relating

A

and

B

and
hopefully we’ll be able to determine the value of the constants.
The easiest values to pick are the roots of the denominator of the
original fraction. In this case, when we plug

x

=

1

in Eq.
(

2

) we get

5

⁢

B

=

2

, and so

B

=

2

/

5

. When we plug

x

=

-

4

we
obtain

-

5

⁢

A

=

2

and so

A

=

-

2

/

5

, and we are done! Now we can finish
the integral:

∫

2

x

2

+

3

⁢

x

-

4

⁢

𝑑

x

=

∫

-

2

/

5

x

+

4

⁢

𝑑

x

+

∫

2

/

5

x

-

1

⁢

𝑑

x

=

-

2

5

⁢

ln

⁡

|

x

+

4

|

+

2

5

⁢

ln

⁡

|

x

-

1

|

+

C

=

2

5

⁢

ln

⁡

|

x

-

1

x

+

4

|

+

C

.

The method.

The goal of the method, as explained above, is
to express any fraction

p

⁢

(

x

)

q

⁢

(

x

)

as the sum of

partial fractions

of the types discussed in the previous section.

1.

If the degree of

p

⁢

(

x

)

is larger than the degree of

q

⁢

(

x

)

then use polynomial

division

to divide and obtain a

quotient

t

⁢

(

x

)

and

remainder

r

⁢

(

x

)

polynomials such that

p

⁢

(

x

)

=

q

⁢

(

x

)

⁢

t

⁢

(

x

)

+

r

⁢

(

x

)

. Thus

p

⁢

(

x

)

q

⁢

(

x

)

=

t

⁢

(

x

)

+

r

⁢

(

x

)

q

⁢

(

x

)

where the degree of

r

is lower than the degree of

q

. Now use
the

partial fraction decomposition

with

r

⁢

(

x

)

/

q

⁢

(

x

)

.

2.

Factor the denominator,

q

⁢

(

x

)

, into

irreducible polynomials

(over

ℝ

). Thus, we may express

q

⁢

(

x

)

as a

product

of linear
polynomials (perhaps to a power other than

1

) and

irreducible

quadratic polynomials

.

3.

If a linear factor

(

x

-

a

)

to the

first

power appears
in the denominator of

q

⁢

(

x

)

, the partial fraction decomposition
should have a term

A

(

x

-

a

)

, for some constant

A

to be
determined.

4.

If a linear factor to the

nth

power, say

(

x

-

b

)

n

appears in the denominator of

q

⁢

(

x

)

, the partial fraction
decomposition should have terms

B

1

(

x

-

b

)

+

B

2

(

x

-

b

)

2

+

…

+

B

n

(

x

-

b

)

n

for some

n

constants

B

1

,

B

2

,

…

,

B

n

to be determined.

5.

If a quadratic polynomial

a

⁢

x

2

+

b

⁢

x

+

c

to the

n

th power appears in the

factorization

of

q

⁢

(

x

)

, i.e.

(

a

⁢

x

2

+

b

⁢

x

+

c

)

n

is a factor of

q

⁢

(

x

)

, then the partial fraction decomposition
should have terms

C

1

⁢

x

+

D

1

a

⁢

x

2

+

b

⁢

x

+

c

+

C

2

⁢

x

+

D

2

(

a

⁢

x

2

+

b

⁢

x

+

c

)

2

+

…

+

C

n

⁢

x

+

D

n

(

a

⁢

x

2

+

b

⁢

x

+

c

)

n

for some constants

C

i

,

D

i

to be determined.

6.

Once you have the sum of all appropriate partial fractions
(see above), group together all these partial fractions into one
fraction

P

⁢

(

x

)

q

⁢

(

x

)

(use a

minimum

common

multiple

for
the denominator! The minimum common multiple will actually be

q

⁢

(

x

)

). In order to have an equality you need to find appropriate
constants

A

,

B

,

…

such that

p

⁢

(

x

)

=

P

⁢

(

x

)

. For this, plug
values of

x

to obtain equations relating the constants.

Example 2.2

.

Suppose we want to find the partial fraction decomposition of:

3

⁢

x

+

2

(

x

-

1

)

⁢

(

x

-

2

)

⁢

(

x

-

3

)

2

⁢

(

1

+

x

2

)

Then, we need constants

A

,

B

,

C

,

D

,

E

,

F

such that

3

⁢

x

+

2

(

x

-

1

)

⁢

(

x

-

2

)

⁢

(

x

-

3

)

2

⁢

(

1

+

x

2

)

=

A

x

-

1

+

B

x

-

2

+

C

x

-

3

+

D

(

x

-

3

)

2

+

E

⁢

x

+

F

1

+

x

2

.

The next step would be to add up all the partial fractions into
one big fraction

P

⁢

(

x

)

(

x

-

1

)

⁢

(

x

-

2

)

⁢

(

x

-

3

)

2

⁢

(

1

+

x

2

)

and
find constants

A

,

B

,

C

,

D

,

E

,

F

such that

P

⁢

(

x

)

=

3

⁢

x

+

2

for all

x

.

Title

a lecture on the partial fraction decomposition method

Canonical name

ALectureOnThePartialFractionDecompositionMethod

Date of creation

2013-03-22 15:39:12

Last modified on

2013-03-22 15:39:12

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

5

Author

alozano (2414)

Entry type

Feature

Classification

msc 26A42

Classification

msc 28-00

Related topic

ALectureOnIntegrationBySubstitution

Related topic

ALectureOnIntegrationByParts

Related topic

ALectureOnTrigonometricIntegralsAndTrigonometricSubstitution

Related topic

PartialFractionsOfExpressions

Related topic

PartialFractionsForPolynomials

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