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{{Math-h}}

'''Integration by partial fractions''' is a [[Techniques of integration|technique]] in [[Calculus]] to facilitate the integration of a rational expression by partial fraction decomposition.

'''Integration by partial fractions''' is a [[Techniques of integration|technique]] in [[Calculus]] to facilitate the [[integration]] of a rational expression by partial fraction decomposition. Given an integral

 

:<math>\int \frac {f(x)}{g(x)}dx</math>

 

where <math>f(x)</math> and <math>g(x)</math> are both [[polynomial]]s, integration by partial fractions shows how to separate the problem into multiple integrals before integrating.

Given an integral

<math>\int \frac {f(x)}{g(x)}dx</math>

<br />

where <math>f(x)</math> and <math>g(x)</math> are both polynomials, integration by partial fractions shows how to separate the problem into multiple integrals before integrating.

==Integration by Partial Fractions==

==Integration by Partial Fractions==

=== A 1st-Degree Denominator ===

=== A 1st-Degree Denominator ===

Given:

Given:

<br /><math>\int \frac{1}{ax + b}dx</math>

:<math>\int \frac{1}{ax + b}dx</math>

<br />Substitute <math>u = ax + b</math>

Substitute <math>u = ax + b</math>

<br /><math>= \int \frac{1}{u} \frac{du}{a} = \frac{1}{a} \int \frac{du}{u} = \frac{1}{a} \ln{|u|} + C = \frac{1}{a} \ln{|ax + b|} + C</math>

:<math>= \int \frac{1}{u} \frac{du}{a} = \frac{1}{a} \int \frac{du}{u} = \frac{1}{a} \ln{|u|} + C = \frac{1}{a} \ln{|ax + b|} + C</math>

This means that if given an integral such as:

This means that if given an integral such as:

<br /><math>\int \frac{8}{3x+13}dx</math>

:<math>\int \frac{8}{3x+13}dx</math>

<br />The steps can be skipped by using the general formula above get:

The steps can be skipped by using the general formula above get:

<br /><math>= \frac{8}{3} \ln{|3x+13|}+C</math>

:<math>= \frac{8}{3} \ln{|3x+13|}+C</math>

=== A Repeated 1st-Degree Denominator ===

=== A Repeated 1st-Degree Denominator ===

The formula for integrals where a first degree polynomial denominator is raised to a power greater than one is much different than the formula above.

The formula for integrals where a first degree polynomial denominator is raised to a power greater than one is much different than the formula above.

<br />Given:

Given:

:<math>\int \frac{1}{(ax+b)^k}dx</math>

:<math>\int \frac{1}{(ax+b)^k}dx</math>

:<math>u = ax+b</math>

:<math>u = ax+b</math>

:<big><math>\frac{3x+11}{(x-3)(x+2)}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}</math></big>

:<big><math>\frac{3x+11}{(x-3)(x+2)}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}</math></big>

Therefore the problem can be restated,

Therefore, the problem can be restated,

:<big><math>3x+11=A(x+2)+B(x-3)</math></big>

:<big><math>3x+11=A(x+2)+B(x-3)</math></big>

:<big><math>\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c</math></big>

:<big><math>\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c</math></big>

==See Also==

==See also==

*[[Calculus]]

*[[Calculus]]

*[[Integral]]

*[[Integral]]

*[[Methods of integration]]

*[[Methods of integration]]

[[Category:calculus]]

[[Category:Calculus]]

[[Category:integration]]

[[Category:Integration]]