Partial Fraction Decompositions Rational Functions Partial Fraction Decompositions Finding Partial Fractions Decompositions Integrating Partial Fraction

Partial Fraction Decompositions

Rational FunctionsPartial Fraction DecompositionsFinding Partial Fractions DecompositionsIntegrating Partial Fraction DecompositionsExamples

Mika Seppälä: Partial Fraction

Decompositions

Rational FunctionsA function of the type P/Q, where both P and Q are polynomials, is a rational function.

Definition

3

2

1 is a rational function.

1

x

x x

Example

The degree of the denominator of the above rational function is less than the degree of the numerator. Hence we may rewrite the above rational function in a simpler form by performing polynomial division.

3

2 2

1 21

1 1

xx

x x x x

Rewriting

For integration, it is always necessary to perform polynomial division first, if possible. To integrate the polynomial part is easy, and one can reduce the problem of integrating a general rational function to a problem of integrating a rational function whose denominator has degree greater than that of the numerator.


Decompositions

Example of Integrating Rational Functions 2 2

2

1 1 2Integrating the functions , and is a simple task

1 1 1applying basic integration formulae and, in the last case, the

substitution 1 .

x

x x x

u x

22 2

One gets here we omit the constants of integration

1 1 2ln 1 , arctan and ln 1

1 1 1

xdx x dx x dx xx x x

22 2

Hence

1 1 2ln 1 arctan ln 1 .

1 1 1

xdx x x x C

x x x

2

23 2

3 3 2I.e. ln 1 arctan ln 1 .

1

x xdx x x x C

x x x


Decompositions

Partial Fraction Decomposition

2

23 2

The integration

3 3 2 ln 1 arctan ln 1

1was based on the decomposition

x xdx x x x C

x x x

2

3 2 2 2

3 3 2 1 1 2

1 1 1 1

x x x

x x x x x x

of the function to be integrated.

Definition The above decomposition of the rational function (3x2+3x+2)/(x3+x2+x+1) is a partial fraction decomposition.

Partial Fraction Decomposition is a rewriting of a rational function into a sum a rational functions with as simple denominators as possible.

General partial fraction decomposition is technically complicated and involves several cases. It all starts with a factorization of the denominator. The type of the partial factor decomposition depends on the type of the factors of the denominator. The different cases will be explained on the following slides.


Decompositions

Simple Linear Factors (1)

1 1 2 2

Consider a rational function of the type

P P

Q

where 0 for all , for , and deg P deg Q .

n n

jij

i j

x x

x a x b a x b a x b

bba j i j n

a a

1 2

1 1 2 2 1 1 2 2

P

for some uniquely defined numbers , 1, , .

n

n n n n

k

x A A A

a x b a x b a x b a x b a x b a x b

A k n

Case I

Partial Fraction Decomposition: Case I


Decompositions

Simple Linear Factors (2)

2 2

0 10 2 ( ) ( ).

1 1 2 1

A B Ax A B x A B

x x A B B

2

2 2Consider the rational function .

1 1 1x x x

2

By the result concerning Case I we can find numbers and such that

2 2 .

x 1 1 1 1 1

A B

A B

x x x x

Example

2 2

Compute these numbers in the following way

2 2 ( 1) ( 1)

x 1 1 1 1 ( 1)( 1) ( 1)( 1)

A B A x B x

x x x x x x x

2

2 1 1So the partial fraction decomposition is .

x 1 1 1x x

To get the equations for A and B we use the fact that two polynomials are the same if and only if their coefficients are the same.


Decompositions

Simple Quadratic Factors (1)

2

PConsider a rational function of the type , deg P deg Q .

Q

Assume that the denominator Q has a quadratic factor .

x

x

x ax bx c

2

2

The quadratic factor of the denominator leads to a term

of the type in the partial fraction decomposition.

ax bx c

Ax B

ax bx c

Case II

Partial Fraction Decomposition: Case II


Decompositions

Simple Quadratic Factors (2)

3 2

2

3 3The rational function has a term of

1 ( 1)( 1)

the type in its partial fraction decomposition. 1

x x x x

Ax B

x x

2

3 2 3 2 2

3 3 ( )( 1) ( 1)

1 1 1 1 ( 1)( 1) ( 1)( 1)

Ax B C Ax B x C x x

x x x x x x x x x x x

2

3 3

3 ( ) ( )

1 1

A C x C B A x C B

x x

Example

3 2

Hence

3 1 2

x 1 1 1

x

x x x

0

0

3

A C

C B A

C B

1

2.

1

A

B

C

To get these equations use the fact that the coefficients of the two numerators must be the same.


Decompositions

Repeated Linear Factors (1)


Q

Assume that the denominator Q has a repeated linear factor , 1. k

x

x

x ax b k

Case III

1 2

2

The repeated linear factor of the denominator leads to terms

of the type in the partial fraction

decomposition.

k

kk

ax b

A A A

ax b ax b ax b

Partial Fraction Decomposition: Case III


Decompositions

Repeated Linear Factors (2)

2

23 2

2 2

2 3 2

4 4 4

1 1 11

1 1 1 1 4 4 4

11 1

x x A B C

x x x x xx

A x x B x C x x x

x x xx x

2 2

2 3 2

2 4 4 4

11 1

A C x B C x A B C x x

x x xx x

4

2 4

4

A C

B C

A B C

2 2

23 2

2

4 4 4 4 4 4The rational function has

1 1 1

a partial fraction decomposition of the type .1 11

x x x x

x x x x x

A B C

x xx

Example

3

2

1

A

B

C

2

23 2

4 4 4 3 2 1We get .

1 1 11

x x

x x x x xx

Equate the coefficients of the numerators.


Decompositions

Repeated Quadratic Factors (1)

2


Q

Assume that the denominator Q has a repeated quadratic factor

, 1. k

x

x

x

ax bx c k

Case IV

2

1 1 2 222 2 2

The repeated quadratic factor of the denominator leads

to terms of the type

in the partial fraction decomposition.

k

k kk

ax bx c

A x B A x B A x B

ax bx c ax bx c ax bx c

Partial Fraction Decomposition: Case IV


Decompositions

Repeated Quadratic Factors (2)

4 2 4 2

25 4 3 2 2

1 1 2 222 2

2 3 2 3 has a partial

2 2 1 1 1

fraction decomposition of the type .1 11

x x x x x x

x x x x x x x

A x B A x B C

x xx

Example

1 1 2 222 2

22 21 1 2 2

22

1 11

1 1 1 1

1 1

A x B A x B C

x xx

A x B x x A x B x C x

x x

1 1 2

4 2

2 25 4 3 2 2 2

Computing in the same way as before one gets 1,

2 3 1 1and 0. Hence .

2 2 1 1 11

A B A C

x x x x xB

x x x x x x xx

This can also be computed by Maple command convert.


Decompositions

Integrating Partial Fraction Decompositions (1)

1. lnA A

dx ax b Kax b a

1

2. , 1.1

l

l

ax bA Adx K l

a lax b

Partial Fraction Decomposition is the main method to integrate rational functions. After a general partial fraction decomposition one has to deal with integrals of the following types. There are four cases. Two first cases are easy.

Here K is the constant of integration.


Decompositions


22

/ 2ln

2

A B Ab aax bc c dx

a ax bx c

2 2 2

2 / 2

2

Ax B A ax b B Ab adx dx dx

ax bx c a ax bx c ax bx c

2

/ 2To compute another rewriting and a substitution is needed.

B Ab adx

ax bx c

The 3rd case leads to two integrals in general. Here we have to compute integrals of the type:

23.

Ax Bdx

ax bx c

We start by the following rewriting

Compute the first integral by a substitution. One gets:


Decompositions


2

2 2

22 / arctan

/ 2 4We get 4

using a substitution.

ax bB Ab a

B Ab a ac bdx Kax bx c ac b

2

2 22

/ 2To compute rewrite it as follows

/ 2 1.

2 / 2 / 2

B Ab adx

ax bx cB Ab a Ab

dx B dxax bx c a x b a c b a

22Observe that since cannot be factored further, / 2 0.ax bx c c b a


Decompositions


2Integrals of the type , 1, are the last case.

l

Ax Bdx l

ax bx c

All these integrals can also be computed using the previous methods.

PAll rational functions f can be integrated by Partial

QFraction Decompositions provided that the polynomial Q

can be factored.

Theorem


Decompositions

Integration AlgorithmIntegration of a rational function f = P/Q, where P and Q are polynomials

can be performed as follows.

1. If deg(Q) deg(P), perform polynomial division and write P/Q = S + R/Q, where S and R are polynomials with deg(R) < deg(Q).Integrate the polynomial S.

2. Factorize the polynomials Q and R. Cancel the common factors. Perform Partial Fraction Decomposition to the simplified form of R/Q.

3. Integrate the Partial Fraction Decomposition.


Decompositions

Examples (1)

3 23

2

3 2 3 2 2

2

3 3

3. Observe 1 ( 1)( 1).

1

3 3 ( 1) ( )( 1)

1 1 1 1 ( 1)( 1) ( 1)( 1)

0 13 ( ) ( )

0 1.1 1

3 2

Hence

x x x xx

A Bx C A x x Bx C x

x x x x x x x x x x x

A B AA B x A B C x A C

A B C Bx x

A C C

3 2

3 1 2

x 1 1 1

x

x x x


Decompositions

Examples (2)

3

2 2

3

2

2

2

2 2 1 1

1 1 1 1

1 2 1 1

1 1 1

ln | 1| ln | 1|2

1ln .

2 1

x xx x

x x x x

xdx x dx

x x x

xx x C

x xC

x


Decompositions

Examples (3)

3 2

3 2

2

2 2

2

3 1 2

x 1 1 13 1 2

1 1 1

2ln 1

11 2 1 3 1

ln 12 1 2 11 2 1

ln 1 ln 1 3 arctan( )2 3

x

x x xx

dx dxx x x x

xx dx

x xx

x dx dxx x x x

xx x x C

The last two integrals have to be computed by suitable substitutions.

Documents

Partial Fraction Decompositions Rational Functions Partial Fraction Decompositions Finding Partial Fractions Decompositions Integrating Partial Fraction