# What is int x^4/(x^4-1)dx?

Nov 6, 2015

$\left[x\right] - \frac{1}{2} \left[\arctan \left(x\right)\right] - \frac{1}{4} \left[\ln | x + 1 |\right] + \frac{1}{4} \left[\ln | x - 1 |\right] + C$

#### Explanation:

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

Looking at $\int \frac{1}{{x}^{4} - 1} \mathrm{dx}$

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

So

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

You need to do partial fraction, you get

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

$\left[x\right] - \frac{1}{2} \left[\arctan \left(x\right)\right] - \frac{1}{4} \left[\ln | x + 1 |\right] + \frac{1}{4} \left[\ln | x - 1 |\right] + C$