Method of Integration by Partial Fractions

THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS

Page 1

This method is based on the simple concept of adding fractions by getting a common denominator. For

example,

so that we can now say that a partial fractions decomposition for is

This concept can also be used with functions of . For example,

(

)

(

)

( )

( )

so that we can now say that a partial fractions decomposition for

is

First we review a few terms. The most general polynomial is an equation of the form

The are called the coefficients and the largest exponent is called the order of the polynomial.

What we would like to be able to do is find a partial fractions decomposition for a given function. For

example, what would be a partial fractions decomposition for

? Begin by factoring the

denominator, getting

( )( )

Now assume that there are constants and so that

( )( )


Page 2

It can be shown that such constants always exist for the rational function ( ) ( )⁄ , ( ) , if the

both ( ) and ( ) are polynomials and the degree of ( ) is smaller than the degree of ( ).

The following general rules apply for the method of partial fractions

1. If the degree of ( ) is equal to or greater than the degree of ( ), use polynomial long division in

order to rewrite the given rational function as the sum of a polynomial and a remainder (a new

rational function with ( ) having larger degree than ( )).

2. Factor the denominator as much as possible. Assume the final form looks like

( ) ( )

where cannot be further factored (for example, ).

3. Repeated factors ( ) in the denominator result in terms where is the order of the

factor. For example, if after factoring there is ( ) in the denominator there will be 3 terms

resulting

( )

( )

4. Terms in the denominator that cannot be further factored will generate a term that

has a numerator of one degree lower. So, for example, if the form is (degree 2

polynomial) the numerator for that term would be (degree 1 polynomial).

The following example illustrates the partial fractions decomposition of a rational function, where the

linear factor is repeated three times and the irreducible quadratic factor is repeated twice.

( ) ( )

( )

( )

( )

Example 1

Getting a common denominator and adding the fractions.

( )( )

(

)

(

)

( )

( )( )

( )

( )( )


Page 3

Since the fractions in the above equation have the same denominators, it follows that their numerators

must be equal. Thus,

( ) ( )

The right-hand side of this equation can be considered a function of which is equal to 6 for all values of

. Grouping like coefficients and recognizing that is a quadratic of the form we obtain

( ) ( )

This yields two equations and two unknowns, namely

Note that in the most general case we would have equations and unknowns, where is the order of

( ) in the denominator. These are very easy to solve by inspection. The first is and

substituting into the second yields ( ) , so and

After getting familiar with this process, in order to save some time, get in the habit of going from the

Equation 1 directly to Equation 2 by recognizing that we can "cross-multiply" the terms on the right to

determine the numerators.

Example 2

Factoring the denominator and applying our rules

( )

Cross multiplying on the right to get the numerator term:

( ) ( )

( )

Comparing the coefficients of same powers of on both sides we get


Page 4

Substituting in for A in the first equation yields

and so

Example 3

Factoring the denominator and applying rules

( )( )

Cross multiplying right-hand side for numerator

( ) ( )

Combining like terms

( ) ( )

Which means

This can be solved by substitution or by subtracting 4x the first equation from the second to yield

So

,

and we obtain

( )( )

( )

( )

NOTE: Wolfram has a partial fraction calculator online so you can check your homework answers:

http://www.wolframalpha.com/widgets/view.jsp?id=ec4a062bb304f88c2ba0b631d7acabbc

REFERENCE: Information was taken from "Method of integration by partial fractions"

https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/

http://www.wolframalpha.com/widgets/view.jsp?id=ec4a062bb304f88c2ba0b631d7acabbc

https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/partialfracdirectory/


Page 5

Example 4

( ) ( )

This is already factored so applying our rules

( ) ( )

( )

( )

( )

A simple "cross-multiplying method" won't work here. Instead recognizing that you need to

multiply the numerators on the right by a term that yields the denominator on the left, we get

( ) ( ) ( )( ) ( )

( )( ) ( ) ( )( )

Expanding and combining like terms

( )( )

( )( )

( )

( )( )( )( )

( )( )( )

( )

( ) ( )

( )( )( )

( )( )

( )

( )

( )

( )( )

( ) ( )

( )

( )

( )

( )

( )

( )

( )


Page 6

Combining like terms

( )

( )

( )

( )

( )

( )

( )

Equating terms we obtain 7 equations and 7 unknowns

[ ]

[ ]

[

]

Which in augmented matrix form and on reducing

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Now do back substitution

[

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⁄

]

[

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⁄

⁄ ]

[

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⁄

⁄ ]

[

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⁄

⁄

⁄ ]

[

]

[ ]

[

⁄

⁄

⁄

]


Page 9

( ) ( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

∫

( ) ( )

∫

( ) ∫

( )

∫

( )

∫

( ) ∫

( )

( )

( )

( ) ∫

( )

Let ,

∫

( ) ∫

( )

∫

( )

∫

∫

∫( )

√ ⁄ √ ⁄ so

∫

( )

( )

And so

∫

( ) ( )

( )

( )

( ) (

)


Page 10

EXTRA CREDIT (20 pts): Integrate the following rational expression. Requires: partial fraction

decomposition, reducing an 8x8 matrix to row-echelon form and back substituting, and trigonometric

substitution to solve two of the resulting integrals.

( ) ( )

( )

( )

( )

( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( )

( ) ( )

( )( )

( )( ) ( )( ) ( )( )( ) ( )( )( ) ( )( )

( ( ) ( )( ) ( )( ) ( )( )

( ( ) ( )( ) ( )( ) ( )( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )


Page 11

(

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(

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(

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(

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(

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(

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(

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(

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)


Page 12

(

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(

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)

Now back substitution

(

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⁄

⁄

⁄

)

(

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⁄

⁄

⁄

⁄

)

(

|

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⁄

⁄

⁄

⁄

⁄

)

(

|

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⁄

⁄

⁄

⁄

⁄

⁄

)

(

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⁄

⁄

⁄

⁄

⁄

⁄

⁄

)

(

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⁄

⁄

⁄

⁄

⁄

⁄

⁄

⁄

)


Page 13

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

Integrating

∫

∫

( )

∫

∫

∫

( )

∫

( )

∫

( )

∫

( )

| |

( )

( )

(

)

( )

∫

( )

( )

∫

( )

Last two integrals by trigonometric substitution , .

∫

( )

∫

( )

∫

( )

∫

( )

∫

( )

∫

( )

∫

( )

∫

( )

∫

( )

∫

∫( )

∫

∫(

)

∫

( )

∫( )( )

∫( )

∫( )

(

)

∫

∫(

) ( )

∫


Page 14

( )

( )

( )( )

( )

Substituting back, ⁄ ⁄ ,

√ ,

√ ,

(

)

(

) (

)

(

)

( )

( )

Combining with previous result we get

| |

( )

( )

(

)

( )

∫

( )

( )

∫

( )

| |

( )

( )

(

)

( )

( )

(

)

( )

( )

| |

( )

( )

(

)

(

)

( )

( )

( )

( )

| |

( )

( )

(

)

( )

( )


Page 15

| |

( )

( )

(

)

( )

( )

FINAL ANSWER:

∫

( ) ( )

[ | | ( ) (

)

( )

( )

( )

( ) ]

Method of Integration by Partial Fractions

Documents

method of partial fractions

partial fractions page

common denominator

following example

denominator result

larger degree

degree lower

general polynomial