Partial fractions in integration

${\frac {d}{dx}}\sin x=?\,$

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

$\int {\frac {f(x)}{g(x)}}dx$

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.

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

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

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

See also

• Calculus
• Integral
• Methods of integration