# How do you evaluate the integral int x^4/(x^2-1)dx?

Mar 4, 2018

$\int {x}^{4} / \left({x}^{2} - 1\right) \mathrm{dx} = \frac{1}{3} {x}^{3} + x + \frac{1}{2} \ln \left\mid \frac{x - 1}{x + 1} \right\mid + \text{c}$

#### Explanation:

For the integrand ${x}^{4} / \left({x}^{2} - 1\right)$, we perform a long division to get it into an integrable form.

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

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

We now perform a partial fraction decomposition on the 2nd term

$\frac{1}{\left(x + 1\right) \left(x - 1\right)} = \frac{\left(x + 1\right) - \left(x - 1\right)}{2 \left(x + 1\right) \left(x - 1\right)} =$

$\frac{x + 1}{2 \left(x + 1\right) \left(x - 1\right)} - \frac{x - 1}{2 \left(x + 1\right) \left(x - 1\right)} = \frac{1}{2 \left(x - 1\right)} - \frac{1}{2 \left(x + 1\right)}$

So

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

and we can now integrate:

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

$= \frac{1}{3} {x}^{3} + x + \frac{1}{2} \ln \left\mid x - 1 \right\mid - \frac{1}{2} \ln \left\mid x + 1 \right\mid + \text{c}$

$= \frac{1}{3} {x}^{3} + x + \frac{1}{2} \ln \left\mid \frac{x - 1}{x + 1} \right\mid + \text{c}$