Partial fractions in integration

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

Given an integral
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac {f(x)}{g(x)}dx}
where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} and $g(x)$ are both polynomials, integration by partial fractions shows how to separate the problem into multiple integrals before integrating.

1st-Degree Denominators

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

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:
$={\frac {8}{3}}\ln {|3x+13|}+C$

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
$\int {\frac {1}{(ax+b)^{k}}}dx$
$u=ax+b$
$=\int {\frac {1}{u^{k}}}{\frac {du}{a}}={\frac {1}{a}}\int u^{-k}du={\frac {1}{a}}\cdot {\frac {u^{k-1}}{-k}}+C={-1 \over kau^{k-1}}+C={-1 \over {ka(ax+b)^{k-1}}}+C$
This means that integrals such as
$\int {5 \over {(3x+17)^{12}}}dx$
are now very easy:
$={-5 \over {(3)(12)(3x+17)^{11}}}+C={-5 \over {36(3x+17)^{11}}}+C$

2nd-Degree Denominators

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

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,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-2} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3(-2)+11=A(-2+2)+B(-2-3)}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 5=A(0)+B(5)}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B=-1}

Let :Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=3} ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3(3)+11=A(3+2)+B(3-3)}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 20=A(5)+B(0)}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=4}

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{4}{x-3}-\frac{1}{x+2}}

Now solve the integral.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}-\frac{1}{x+2}dx}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}dx-\int\frac{1}{x+2}dx}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c}