# How do you integrate int (x^4)/(x^(5)+1) ?

May 5, 2015

$\int \frac{{x}^{4} \mathrm{dx}}{{x}^{5} + 1} = \frac{1}{5} \ln \left({x}^{5} + 1\right) + C$

Method

$\int \frac{{x}^{4} \mathrm{dx}}{{x}^{5} + 1} = \frac{1}{5} \int \frac{5 {x}^{4} \mathrm{dx}}{{x}^{5} + 1} = \frac{1}{5} \int \frac{d \left({x}^{5} + 1\right)}{{x}^{5} + 1} = \text{ I}$

NB: $d \left({x}^{5} + 1\right)$ simply means that the derivative of ${x}^{5} + 1$ is $5 {x}^{4}$

Now, you could as well let $u = {x}^{5} + 1$

So that,

$\text{I} = \frac{1}{5} \int \frac{\mathrm{du}}{u} = \frac{1}{5} \ln u = \frac{1}{5} \ln \left({x}^{5} + 1\right) + C$