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

Aug 5, 2016

$= - \frac{1}{3} \left({x}^{2} + 2\right) \sqrt{1 - {x}^{2}} \setminus + C$

#### Explanation:

$\int \setminus {x}^{3} / \sqrt{1 - {x}^{2}} \setminus \mathrm{dx}$

$= \int \setminus {x}^{2} \frac{d}{\mathrm{dx}} \left(- \sqrt{1 - {x}^{2}}\right) \setminus \mathrm{dx}$

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

$= - {x}^{2} \sqrt{1 - {x}^{2}} + \int \setminus 2 x \sqrt{1 - {x}^{2}} \setminus \mathrm{dx}$

$= - {x}^{2} \sqrt{1 - {x}^{2}} + \int \setminus \frac{d}{\mathrm{dx}} \left(- \frac{2}{3} {\left(1 - {x}^{2}\right)}^{\frac{3}{2}}\right) \setminus \mathrm{dx}$

$= - {x}^{2} \sqrt{1 - {x}^{2}} - \frac{2}{3} {\left(1 - {x}^{2}\right)}^{\frac{3}{2}} + C$

$= \sqrt{1 - {x}^{2}} \left(- {x}^{2} - \frac{2}{3} \left(1 - {x}^{2}\right)\right) + C$

$= \sqrt{1 - {x}^{2}} \left(- {x}^{2} / 3 - \frac{2}{3}\right) + C$

$= - \frac{1}{3} \left({x}^{2} + 2\right) \sqrt{1 - {x}^{2}} \setminus + C$