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Partial fractions in integration (view source)
Revision as of 13:49, December 30, 2008
, 13:49, December 30, 2008Created the Partial fractions in integration page. See talk page for possible future edits.
'''Integration by partial fractions''' is a [[Techniques of integration|technique]] to facilitate the integration of a rational expression by partial fraction decomposition.
Given an integral
<br />
<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==
== 1st-Degree Denominators ==
These are a few methods of solving integrals with first degree denominators.
=== 1st-Degree Denominator ===
Given:
<br /><math>\int \frac{1}{ax + b}dx</math>
<br />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>
This means that if given an integral such as:
<br /><math>\int \frac{8}{3x+13}dx</math>
<br />The steps can be skipped by using the general formula above get:
<br /><math>= \frac{8}{3} \ln{|3x+13|}+C</math>
=== 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.
Given
<br /><math>\int \frac{1}{(ax+b)^k}dx</math>
<br /><math>u = ax+b</math>
<br /><math>= \int \frac{1}{u^k} \frac{du}{a} = \frac{1}{a} \int u^{-k} du = \frac{1}{a} \cdot \frac{u^{k-1}}{-k} + C = {-1 \over kau^{k-1}} + C = {-1 \over {ka (ax+b)^{k-1}}} + C</math>
<br />This means that integrals such as
<br /><math>\int {5 \over {(3x+17)^{12}}}dx</math>
<br />are now very easy:
<br /><math>= {-5 \over {(3)(12)(3x+17)^{11}}}+C = {-5 \over {36(3x+17)^{11}}}+C</math>
== 2nd-Degree Denominators ==
2nd-Degree Denominators get more complicated, especially with those that do not factor.
===Reducible 2nd-Degree Polynomial Denominator===
:<big><math>\int\frac{3x+11}{x^2-x-6}dx</math></big>
The first step is to factor the denominator as much as possible and get the form of the partial fraction decomposition. Doing this gives,
:<big><math>\frac{3x+11}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}</math></big>
This allows the denominator to split the fraction in to sums by cross multiplying the denominators,
:<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,
:<big><math>3x+11=A(x+2)+B(x-3)</math></big>
Now it is possible to solve for A and B by substituting x with a value that allows the term to go to 0. For example,
Let :<big><math>x=-2</math></big>,
:<big><math>3(-2)+11=A(-2+2)+B(-2-3)</math></big>
:<big><math>5=A(0)+B(5)</math></big>
:<big><math>B=-1</math></big>
Let :<big><math>x=3</math></big>,
:<big><math>3(3)+11=A(3+2)+B(3-3)</math></big>
:<big><math>20=A(5)+B(0)</math></big>
:<big><math>A=4</math></big>
Then plug in the values of A and B and get,
:<big><math>\frac{4}{x-3}-\frac{1}{x+2}</math></big>
Now solve the integral.
:<big><math>\int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}-\frac{1}{x+2}dx</math></big>
:<big><math>\int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}dx-\int\frac{1}{x+2}dx</math></big>
:<big><math>\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c</math></big>
Given an integral
<br />
<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==
== 1st-Degree Denominators ==
These are a few methods of solving integrals with first degree denominators.
=== 1st-Degree Denominator ===
Given:
<br /><math>\int \frac{1}{ax + b}dx</math>
<br />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>
This means that if given an integral such as:
<br /><math>\int \frac{8}{3x+13}dx</math>
<br />The steps can be skipped by using the general formula above get:
<br /><math>= \frac{8}{3} \ln{|3x+13|}+C</math>
=== 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.
Given
<br /><math>\int \frac{1}{(ax+b)^k}dx</math>
<br /><math>u = ax+b</math>
<br /><math>= \int \frac{1}{u^k} \frac{du}{a} = \frac{1}{a} \int u^{-k} du = \frac{1}{a} \cdot \frac{u^{k-1}}{-k} + C = {-1 \over kau^{k-1}} + C = {-1 \over {ka (ax+b)^{k-1}}} + C</math>
<br />This means that integrals such as
<br /><math>\int {5 \over {(3x+17)^{12}}}dx</math>
<br />are now very easy:
<br /><math>= {-5 \over {(3)(12)(3x+17)^{11}}}+C = {-5 \over {36(3x+17)^{11}}}+C</math>
== 2nd-Degree Denominators ==
2nd-Degree Denominators get more complicated, especially with those that do not factor.
===Reducible 2nd-Degree Polynomial Denominator===
:<big><math>\int\frac{3x+11}{x^2-x-6}dx</math></big>
The first step is to factor the denominator as much as possible and get the form of the partial fraction decomposition. Doing this gives,
:<big><math>\frac{3x+11}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}</math></big>
This allows the denominator to split the fraction in to sums by cross multiplying the denominators,
:<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,
:<big><math>3x+11=A(x+2)+B(x-3)</math></big>
Now it is possible to solve for A and B by substituting x with a value that allows the term to go to 0. For example,
Let :<big><math>x=-2</math></big>,
:<big><math>3(-2)+11=A(-2+2)+B(-2-3)</math></big>
:<big><math>5=A(0)+B(5)</math></big>
:<big><math>B=-1</math></big>
Let :<big><math>x=3</math></big>,
:<big><math>3(3)+11=A(3+2)+B(3-3)</math></big>
:<big><math>20=A(5)+B(0)</math></big>
:<big><math>A=4</math></big>
Then plug in the values of A and B and get,
:<big><math>\frac{4}{x-3}-\frac{1}{x+2}</math></big>
Now solve the integral.
:<big><math>\int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}-\frac{1}{x+2}dx</math></big>
:<big><math>\int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}dx-\int\frac{1}{x+2}dx</math></big>
:<big><math>\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c</math></big>