# Partial fractions in integration

 $\frac{d}{dx} \sin x=?\,$ This article/section deals with mathematical concepts appropriate for a student in late high school or early university.

Integration by partial fractions is a technique in Calculus to facilitate the integration of a rational expression by partial fraction decomposition.

Given an integral
$\int \frac {f(x)}{g(x)}dx$
where f(x) and g(x) are both polynomials, integration by partial fractions shows how to separate the problem into multiple integrals before integrating.

## A 1st-Degree Denominator

These are a few methods of solving integrals with first degree denominators.

### A 1st-Degree Denominator

Given:
$\int \frac{1}{ax + b}dx$
Substitute u = ax + b
$= \int \frac{1}{u} \frac{du}{a} = \frac{1}{a} \int \frac{du}{u} = \frac{1}{a} \ln{|u|} + C = \frac{1}{a} \ln{|ax + b|} + C$

This means that if given an integral such as:
$\int \frac{8}{3x+13}dx$
The steps can be skipped by using the general formula above get:
$= \frac{8}{3} \ln{|3x+13|}+C$

### 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:

$\int \frac{1}{(ax+b)^k}dx$
u = ax + b
$= \int \frac{1}{u^k} \frac{du}{a} = \frac{1}{a} \int u^{-k)} du = \frac{1}{a} \cdot \frac{u^{k-1}}{-(k-1)} + C = {-1 \over (k-1)au^{k-1}} + C = {-1 \over {(k-1)a (ax+b)^{k-1}}} + C$

Note that the above formula only works if $k \neq 1$.
This means that integrals such as

$\int {5 \over {(3x+17)^{12}}}dx$

are now very easy:

$= {-5 \over {(3)(11)(3x+17)^{11}}}+C = {-5 \over {33(3x+17)^{11}}}+C$

## A 2nd-Degree Denominator

2nd-Degree Denominators get more complicated, especially with those that do not factor.

### A Reducible 2nd-Degree Polynomial Denominator

$\int\frac{3x+11}{x^2-x-6}dx$

The first step is to factor the denominator as much as possible and get the form of the partial fraction decomposition. Doing this gives,

$\frac{3x+11}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}$

This allows the denominator to split the fraction in to sums by cross multiplying the denominators,

$\frac{3x+11}{(x-3)(x+2)}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}$

Therefore the problem can be restated,

3x + 11 = A(x + 2) + B(x − 3)

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 :x = − 2,

3( − 2) + 11 = A( − 2 + 2) + B( − 2 − 3)
5 = A(0) + B(5)
B = − 1

Let :x = 3,

3(3) + 11 = A(3 + 2) + B(3 − 3)
20 = A(5) + B(0)
A = 4

Then plug in the values of A and B and get,

$\frac{4}{x-3}-\frac{1}{x+2}$

Now solve the integral.

$\int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}-\frac{1}{x+2}dx$
$\int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}dx-\int\frac{1}{x+2}dx$
$\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c$