# How do you integrate int x^3/(x^2 -1) using partial fractions?

Jun 24, 2016

$= {x}^{2} / 2 + \frac{1}{2} \ln \left({x}^{2} - 1\right) + C$

#### Explanation:

I'm not sure you can do that

from some simple long division, turns out that ${x}^{3} / \left({x}^{2} - 1\right) = x + \frac{x}{{x}^{2} - 1}$

so the integration is

$\int \setminus x + \frac{x}{{x}^{2} - 1} \setminus \mathrm{dx}$ which is straightaway do-able

$\int \setminus x + \textcolor{red}{\frac{x}{{x}^{2} - 1}} \setminus \mathrm{dx}$
$= {x}^{2} / 2 + \frac{1}{2} \ln \left({x}^{2} - 1\right) + C$

note the pattern in the red term of integrand, it is in form

$\frac{\setminus \alpha f ' \left(x\right)}{f \left(x\right)}$ and here $\alpha = \frac{1}{2}$, so you can "just do it"