Method of Integration by Partial Fractions
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Transcript of 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
( )( )
THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS
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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.
( )( )
(
)
(
)
( )
( )( )
( )
( )( )
THE METHOD OF INTEGRATION BY PARTIAL 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
THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS
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/
THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS
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
( )( )
( )( )
( )
( )( )( )( )
( )( )( )
( )
( ) ( )
( )( )( )
( )( )
( )
( )
( )
( )( )
( ) ( )
( )
( )
( )
( )
( )
( )
( )
THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS
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Combining like terms
( )
( )
( )
( )
( )
( )
( )
Equating terms we obtain 7 equations and 7 unknowns
[ ]
[ ]
[
]
Which in augmented matrix form and on reducing
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THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS
Page 7
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THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS
Page 8
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Now do back substitution
[
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⁄
]
[
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⁄
⁄ ]
[
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⁄
⁄ ]
[
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⁄
⁄
⁄ ]
[
]
[ ]
[
⁄
⁄
⁄
]
THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS
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( ) ( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
∫
( ) ( )
∫
( ) ∫
( )
∫
( )
∫
( ) ∫
( )
( )
( )
( ) ∫
( )
Let ,
∫
( ) ∫
( )
∫
( )
∫
∫
∫( )
√ ⁄ √ ⁄ so
∫
( )
( )
And so
∫
( ) ( )
( )
( )
( ) (
)
THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS
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.
( ) ( )
( )
( )
( )
( )( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( )
( ) ( )
( )( )
( )( ) ( )( ) ( )( )( ) ( )( )( ) ( )( )
( ( ) ( )( ) ( )( ) ( )( )
( ( ) ( )( ) ( )( ) ( )( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS
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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)
(
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(
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)
THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS
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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⁄
⁄
⁄
⁄
⁄
⁄
⁄
⁄
)
THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS
Page 13
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
Integrating
∫
∫
( )
∫
∫
∫
( )
∫
( )
∫
( )
∫
( )
| |
( )
( )
(
)
( )
∫
( )
( )
∫
( )
Last two integrals by trigonometric substitution , .
∫
( )
∫
( )
∫
( )
∫
( )
∫
( )
∫
( )
∫
( )
∫
( )
∫
( )
∫
∫( )
∫
∫(
)
∫
( )
∫( )( )
∫( )
∫( )
(
)
∫
∫(
) ( )
∫
THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS
Page 14
( )
( )
( )( )
( )
Substituting back, ⁄ ⁄ ,
√ ,
√ ,
(
)
(
) (
)
(
)
( )
( )
Combining with previous result we get
| |
( )
( )
(
)
( )
∫
( )
( )
∫
( )
| |
( )
( )
(
)
( )
( )
(
)
( )
( )
| |
( )
( )
(
)
(
)
( )
( )
( )
( )
| |
( )
( )
(
)
( )
( )
THE METHOD OF INTEGRATION BY PARTIAL FRACTIONS
Page 15
| |
( )
( )
(
)
( )
( )
FINAL ANSWER:
∫
( ) ( )
[ | | ( ) (
)
( )
( )
( )
( ) ]