# How do you evaluate the integral int dx/(x^4-16)?

Mar 6, 2018

$\frac{1}{32} \ln | \frac{x - 2}{x + 2} | - \frac{1}{16} a r c \tan \left(\frac{x}{2}\right) + C$.

#### Explanation:

Let, $I = \int \frac{\mathrm{dx}}{{x}^{4} - 16} = \int \frac{\mathrm{dx}}{\left({x}^{2} + 4\right) \left({x}^{2} - 4\right)}$

$= \frac{1}{8} \int \frac{8}{\left({x}^{2} + 4\right) \left({x}^{2} - 4\right)} \mathrm{dx}$,

$= \frac{1}{8} \int \frac{\left({x}^{2} + 4\right) - \left({x}^{2} - 4\right)}{\left({x}^{2} + 4\right) \left({x}^{2} - 4\right)} \mathrm{dx}$,

$= \frac{1}{8} \int \left\{\frac{{x}^{2} + 4}{\left({x}^{2} + 4\right) \left({x}^{2} - 4\right)} - \frac{{x}^{2} - 4}{\left({x}^{2} + 4\right) \left({x}^{2} - 4\right)}\right\} \mathrm{dx}$,

$= \frac{1}{8} \int \left\{\frac{1}{{x}^{2} - 4} - \frac{1}{{x}^{2} + 4}\right\} \mathrm{dx}$,

$= \frac{1}{8} \left\{\int \frac{1}{{x}^{2} - {2}^{2}} - \int \frac{1}{{x}^{2} + {2}^{2}} \mathrm{dx}\right\}$,

$= \frac{1}{8} \left\{\frac{1}{2 \times 2} \ln | \frac{x - 2}{x + 2} | - \frac{1}{2} a r c \tan \left(\frac{x}{2}\right)\right\}$,

$\Rightarrow I = \frac{1}{32} \ln | \frac{x - 2}{x + 2} | - \frac{1}{16} a r c \tan \left(\frac{x}{2}\right) + C$.