Partial fractions in integration
| <math>\frac{d}{dx} \sin x=?\,</math> | This article/section deals with mathematical concepts appropriate for late high school or early college. |
Integration by partial fractions is a 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 polynomials, integration by partial fractions shows how to separate the problem into multiple integrals before integrating.
Integration by Partial Fractions
A 1st-Degree Denominator
These are a few methods of solving integrals with first degree denominators.
A 1st-Degree Denominator
Given:
- <math>\int \frac{1}{ax + b}dx</math>
Substitute <math>u = ax + b</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:
- <math>\int \frac{8}{3x+13}dx</math>
The steps can be skipped by using the general formula above get:
- <math>= \frac{8}{3} \ln{|3x+13|}+C</math>
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. Given:
- <math>\int \frac{1}{(ax+b)^k}dx</math>
- <math>u = ax+b</math>
- <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-1)} + C = {-1 \over (k-1)au^{k-1}} + C = {-1 \over {(k-1)a (ax+b)^{k-1}}} + C</math>
Note that the above formula only works if <math>k \neq 1</math>.
This means that integrals such as
- <math>\int {5 \over {(3x+17)^{12}}}dx</math>
are now very easy:
- <math>= {-5 \over {(3)(11)(3x+17)^{11}}}+C = {-5 \over {33(3x+17)^{11}}}+C</math>
A 2nd-Degree Denominator
2nd-Degree Denominators get more complicated, especially with those that do not factor.
A Reducible 2nd-Degree Polynomial Denominator
- <math>\int\frac{3x+11}{x^2-x-6}dx</math>
The first step is to factor the denominator as much as possible and get the form of the partial fraction decomposition. Doing this gives,
- <math>\frac{3x+11}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}</math>
This allows the denominator to split the fraction in to sums by cross multiplying the denominators,
- <math>\frac{3x+11}{(x-3)(x+2)}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}</math>
Therefore, the problem can be restated,
- <math>3x+11=A(x+2)+B(x-3)</math>
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 :<math>x=-2</math>,
- <math>3(-2)+11=A(-2+2)+B(-2-3)</math>
- <math>5=A(0)+B(5)</math>
- <math>B=-1</math>
Let :<math>x=3</math>,
- <math>3(3)+11=A(3+2)+B(3-3)</math>
- <math>20=A(5)+B(0)</math>
- <math>A=4</math>
Then plug in the values of A and B and get,
- <math>\frac{4}{x-3}-\frac{1}{x+2}</math>
Now solve the integral.
- <math>\int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}-\frac{1}{x+2}dx</math>
- <math>\int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}dx-\int\frac{1}{x+2}dx</math>
- <math>\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c</math>