June | 2026 | Racquel Sanderson

Integration Using Partial Fraction Decomposition

Partial fraction decomposition is the process of taking a rational function and decomposing it into simpler rational expressions which are easier to integrate.

We only use partial fractions if the rational function is proper. If

$f(x)=\frac{p(x)}{q(x)}$

then the degree of $p(x)$ must be less than the degree of $q(x)$ .

Types of Partial Fraction Decompositions

Factor	Term in Partial Fraction
$ax+b$	$\frac{A}{ax+b}$
$(ax+b)^n$	$\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} +\cdots+ \frac{A_n}{(ax+b)^n}$
Irreducible quadratic $ax^2+bx+c$	$\frac{Ax+B}{ax^2+bx+c}$

Example

$\int \frac{3x+5}{x^2-x-2}\,dx$

Find

Factorise the denominator:

$x^2-x-2=(x-2)(x+1)$

Write

$\frac{3x+5}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$

Multiplying through by

$(x-2)(x+1)$ :

$3x+5=A(x+1)+B(x-2)$

Let $x=-1$ :

$2=-3B$

$B=-\frac23$

Let $x=2$ :

$11=3A$

$A=\frac{11}{3}$

Hence

$\frac{3x+5}{x^2-x-2} = \frac{11}{3(x-2)} - \frac{2}{3(x+1)}$

Integrating:

$\int\frac{3x+5}{x^2-x-2}\,dx = \int \left( \frac{11}{3(x-2)} - \frac{2}{3(x+1)} \right) dx$

$= \frac{11}{3}\ln|x-2| -\frac23\ln|x+1| +C$

Example

Find

$\int\frac{3x+2}{(x-1)^2}\,dx$

Write

$\frac{3x+2}{(x-1)^2} = \frac{A_1}{x-1} + \frac{A_2}{(x-1)^2}$

Multiplying through by $(x-1)^2$ :

$3x+2=A_1(x-1)+A_2$

Let $x=1$ :

$5=A_2$

Let $x=0$ :

$2=-A_1+5$

$A_1=3$

Therefore

$\int\frac{3x+2}{(x-1)^2}\,dx = \int \left( \frac3{x-1} + \frac5{(x-1)^2} \right) dx$

$= 3\ln|x-1| -\frac5{x-1} +C$

Example

Find

$\int \frac{2x-5} {(x-1)(x^2+1)} \,dx$

Write

$\frac{2x-5} {(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$

Multiplying through:

$2x-5 = A(x^2+1) + (Bx+C)(x-1)$

Let $x=1$ :

$-3=2A$

$A=-\frac32$

Let $x=0$ :

$-5=-\frac32-C$

$C=\frac72$

Let $x=-1$ :

$-7=-3+\left(-B+\frac72\right)(-2)$

$-4=2B-7$

$B=\frac32$

Hence

$\int \frac{2x-5} {(x-1)(x^2+1)} \,dx = \frac12 \int \left( -\frac3{x-1} + \frac{3x}{x^2+1} + \frac7{x^2+1} \right) dx$

$= \frac12 \left( -3\ln|x-1| +\frac32\ln(x^2+1) +7\arctan(x) \right) +C$

$= -\frac32\ln|x-1| +\frac34\ln(x^2+1) +\frac72\arctan(x) +C$

$\begin{equation*}z^n=1, z\in\mathbb{C}\end{equation}$

In polar form $1=cis (2\pi)$

Hence,

$\begin{equation*}z^n=cis(2\pi k), k\in\mathbb{Z}\end{equation}$

$\begin{equation*}z^n=cis(\frac{2\pik}{n})\end{equation}$

For example,

$\begin{equation*}z^4=1\end{equation}$

$\begin{equation*}z^4=cis(2\pi k)\end{equation}$

$\begin{equation*}z_k=cis(\frac{2\pi k}{4})=cis(\frac{\pi k}{2})\end{equation}$

$\begin{equation*}z_0=1\end{equation}$

$\begin{equation*}z_1=cis(\frac{\pi}{2})=i\end{equation}$

$\begin{equation*}z_2=cis(\pi)=-1\end{equation}$

$\begin{equation*}z_3=cis(\frac{3\pi}{2})=-i\end{equation}$

The roots of unity are spread evenly $(\frac{2\pi}{n}$ or $\frac{360}{n}$ apart) around a circle of radius $1$ .

Properties of the roots of Unity

All the roots of unity can be generated by powers of a single root $\omega=cis(\frac{2\pi}{n})$ . The roots form the sequence $1, \omega, \omega^2, \omega^3, ..., \omega^{n-1}$ Note: $\omega \ne 1$