# How do you integrate 1 / ((x^2 + 1) (x^2 +4)) using partial fractions?

Sep 8, 2016

$\frac{1}{6} \left\{2 a r c \tan x - a r c \tan \left(\frac{x}{2}\right)\right\} + C$.

#### Explanation:

Let $I = \int \frac{1}{\left({x}^{2} + 1\right) \left({x}^{2} + 4\right)} \mathrm{dx}$.

We will use the Method of Partial Fraction to decompose the

integrand $\frac{1}{\left({x}^{2} + 1\right) \left({x}^{2} + 4\right)} = \frac{1}{\left(y + 1\right) \left(y + 4\right)}$, where, $y = {x}^{2}$.

For this, we have to find $A , B \in \mathbb{R}$ such that,

$\frac{1}{\left(y + 1\right) \left(y + 4\right)} = \frac{A}{y + 1} + \frac{B}{y + 4}$.

Using Heavyside's Cover-up Method, we have,

$A = {\left[\frac{1}{y + 4}\right]}_{y = - 1} = \frac{1}{3}$.

$B = {\left[\frac{1}{y + 1}\right]}_{y = - 4} = - \frac{1}{3}$.

$\therefore \frac{1}{\left(y + 1\right) \left(y + 4\right)} = \frac{\frac{1}{3}}{y + 1} + \frac{- \frac{1}{3}}{y + 4}$. Since, $y = {x}^{2}$,

$\frac{1}{\left({x}^{2} + 1\right) \left({x}^{2} + 4\right)} = \frac{1}{3} \left(\frac{1}{{x}^{2} + 1}\right) - \frac{1}{3} \left(\frac{1}{{x}^{2} + 4}\right)$. Therefore,

$I = \frac{1}{3} \int \left(\frac{1}{{x}^{2} + 1}\right) \mathrm{dx} - \frac{1}{3} \int \left(\frac{1}{{x}^{2} + {2}^{2}}\right) \mathrm{dx}$

$= \frac{1}{3} a r c \tan x - \frac{1}{3} \cdot \frac{1}{2} a r c \tan \left(\frac{x}{2}\right)$, i.e.,

$I = \frac{1}{6} \left\{2 a r c \tan x - a r c \tan \left(\frac{x}{2}\right)\right\} + C$.

Enjoy Maths.!