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

Jun 8, 2015

You can do a u-substitution.

Notice how this can be written as:

$\int {x}^{2} \cdot \left(\frac{1}{1 - {x}^{3}}\right) \mathrm{dx}$

$\frac{d}{\mathrm{dx}} \left[- {x}^{3}\right] = - 3 {x}^{2} \mathrm{dx}$, so $\frac{d}{\mathrm{dx}} \left[\left(- \frac{1}{3}\right) \cdot \left(- {x}^{3}\right)\right] = {x}^{2} \mathrm{dx}$.

Let:
$u = 1 - {x}^{3}$
$\mathrm{du} = - 3 {x}^{2} \mathrm{dx}$

$\implies - \frac{1}{3} \int \left(\frac{1}{1 - {x}^{3}}\right) \cdot \left(- 3 {x}^{2}\right) \mathrm{dx}$

$= - \frac{1}{3} \int \frac{1}{u} \mathrm{du}$

$= - \frac{1}{3} \ln | u | + C$

$= - \frac{1}{3} \ln | 1 - {x}^{3} | + C$