The Method of Partial Fractions

Partial fraction expansions provide an approach to integrating a general rational function. Any rational function of a real variable can be written as the sum of a polynomial and a finite number of rational fractions whose denominator is the power of an irreducible polynomial and whose numerator has a degree lower than the degree of this irreducible polynomial. Here are some common examples.

A 1st-Degree Polynomial in the Denominator

The substitution $u = ax + b$, $du = a \,dx$ reduces the integral $\int {1 \over ax+b}\,dx$ to:

$\begin{aligned}\int {1 \over u}\,{du \over a}&={1 \over a}\int{du\over u}\\ &={1 \over a}\ln\left|u\right|+C \\ &= {1 \over a} \ln\left|ax+b\right|+C\end{aligned}$

A Repeated 1st-Degree Polynomial in the Denominator

The same substitution reduces such integrals as $\int {1 \over (ax+b)^8}\,dx$ to 

$\begin{aligned}\int {1 \over u^8}\,{du \over a}&={1 \over a}\int u^{-8}\,du \\ &= {1 \over a} \cdot{u^{-7} \over(-7)}+C \\ &= {-1 \over 7au^7}+C \\ &= {-1 \over 7a(ax+b)^7}+C\end{aligned}$

An Irreducible 2nd-Degree Polynomial in the Denominator

Next we consider integrals such as 

$\displaystyle{\int {x+6 \over x^2-8x+25}\,dx}$

The quickest way to see that the denominator, $x^2 − 8x + 25$, is irreducible is to observe that its discriminant is negative. Alternatively, we can complete the square:

$\begin{aligned}x^2-8x+25&=(x^2-8x+16)+9\\ &=(x-4)^2+9\end{aligned}$

and observe that this sum of two squares can never be $0$ while $x$ is a real number. In order to make use of the substitution

$\begin{aligned} u & = x^2-8x+25 \\ du & =(2x-8)\,dx \\ \frac{du}{2} & = (x-4)\,dx \end{aligned}$

we would need to find $(x-4)$ in the numerator. So we decompose the numerator $x + 6$ as $(x-4) + 10$, and we write the integral as 

$\displaystyle{\int {x-4 \over x^2-8x+25}\,dx + \int {10 \over x^2-8x+25}\,dx}$

The substitution handles the first summand, thus: 

$\begin{aligned}\int \frac{x-4}{x^2-8x+25}\,dx &= \int \frac{\displaystyle{\frac{du}{2}}}{u} \\ &= \frac{1}{2}\ln\left|u\right|+C \\ &= \frac{1}{2}\ln(x^2-8x+25)+C\end{aligned}$

Note that the reason we can discard the absolute value sign is that, as we observed earlier, $(x-4)^2 + 9$ can never be negative.

Next we must treat the integral 

$\displaystyle{\int {10 \over x^2-8x+25} \, dx}$

With a little more algebra, 

$\begin{aligned} \int {10 \over x^2-8x+25} \, dx &= \int {10 \over (x-4)^2+9} \, dx \\ & = \int \frac{\displaystyle{\frac{10}{9}}}{\left(\displaystyle{\frac{x-4}{3}}\right)^2+1}\,dx \\ &= {10 \over 3} \arctan\left(\frac{x-4}{3}\right) + C \end{aligned}$

Putting it all together:

$\displaystyle{\int {x + 6 \over x^2-8x+25}\,dx = {1 \over 2}\ln(x^2-8x+25) + {10 \over 3} \arctan\left({x-4 \over 3}\right) + C}$