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

Aug 30, 2016

$\frac{1}{5} \ln | x - 1 | - \frac{1}{10} \ln \left({x}^{2} + 4\right) + \frac{2}{5} a r c \tan \left(\frac{x}{2}\right) + K$.

#### Explanation:

To use the Method of Partial Fractions, we let, for, $A , B , C \in \mathbb{R} , :$

$\frac{x}{\left(x - 1\right) \left({x}^{2} + 4\right)} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + 4} \ldots \ldots \ldots \ldots \left(1\right)$.

$A$ can easily be obtained by using Heavyside's Cover-up Method :

$A = {\left[\frac{x}{{x}^{2} + 4}\right]}_{x = 1} = \frac{1}{5}$. Hence, with this $A$, from (1), we have,

$\therefore \frac{x}{\left(x - 1\right) \left({x}^{2} + 4\right)} - \frac{\frac{1}{5}}{x - 1} = \frac{B x + C}{{x}^{2} + 4}$, i.e.,

$\frac{x - \frac{1}{5} \left({x}^{2} + 4\right)}{\left(x - 1\right) \left({x}^{2} + 4\right)} = \frac{B x + C}{{x}^{2} + 4}$

$\therefore \frac{\cancel{\left(x - 1\right)} \left(- \frac{1}{5} \cdot x + \frac{4}{5}\right)}{\cancel{\left(x - 1\right)} \left({x}^{2} + 4\right)} = \frac{B x + C}{{x}^{2} + 4}$.

$\Rightarrow B = - \frac{1}{5} , C = \frac{4}{5}$. Hence, by $\left(1\right)$, we have,

$\int \frac{x}{\left(x - 1\right) \left({x}^{2} + 4\right)} \mathrm{dx} = \frac{1}{5} \int \frac{1}{x - 1} \mathrm{dx} + \int \frac{\left(- \frac{1}{5} \cdot x + \frac{4}{5}\right)}{{x}^{2} + 4} \mathrm{dx}$

$= \frac{1}{5} \ln | x - 1 | - \frac{1}{10} \int \frac{2 x}{{x}^{2} + 4} \mathrm{dx} + \frac{4}{5} \int \frac{1}{{x}^{2} + 4} \mathrm{dx}$

$= \frac{1}{5} \ln | x - 1 | - \frac{1}{10} \int \frac{d \left({x}^{2} + 4\right)}{{x}^{2} + 4} + \frac{4}{5} \cdot \left(\frac{1}{2}\right) a r c \tan \left(\frac{x}{2}\right)$

$= \frac{1}{5} \ln | x - 1 | - \frac{1}{10} \ln \left({x}^{2} + 4\right) + \frac{2}{5} a r c \tan \left(\frac{x}{2}\right) + K$.

Enjoy Maths.!

Note : The Method used to find $B , \mathmr{and} , C$ above, the following page from Wikipedia was used :

http://www.math-cs.gordon.edu/courses/ma225/handouts/heavyside.pdf