# How do you integrate x^3 / (4-x^2)?

Apr 11, 2017

$- \frac{1}{2} {x}^{2} - 2 \ln \left\mid 4 - {x}^{2} \right\mid + C$

#### Explanation:

We can rewrite the original function:

${x}^{3} / \left(4 - {x}^{2}\right) = \frac{- x \left(4 - {x}^{2}\right) + 4 x}{4 - {x}^{2}}$

$\textcolor{w h i t e}{{x}^{3} / \left(4 - {x}^{2}\right)} = - x + \frac{4 x}{4 - {x}^{2}}$

If you're uncomfortable with this method of simplification, you can also perform the long division for ${x}^{3} / \left(4 - {x}^{2}\right)$ to find that these are equivalent expressions.

Then:

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

The first integral is simple:

$= - \frac{1}{2} {x}^{2} + 4 \int \frac{x}{4 - {x}^{2}} \mathrm{dx}$

For the second integral, try the substitution $u = 4 - {x}^{2}$. This implies that $\mathrm{du} = - 2 x \mathrm{dx}$. We have the derivative off by a factor of $- 2$ already in the numerator:

$= - \frac{1}{2} {x}^{2} - 2 \int \frac{- 2 x}{4 - {x}^{2}} \mathrm{dx}$

$= - \frac{1}{2} {x}^{2} - 2 \int \frac{1}{u} \mathrm{du}$

This is a common integral:

$= - \frac{1}{2} {x}^{2} - 2 \ln \left\mid u \right\mid$

$= - \frac{1}{2} {x}^{2} - 2 \ln \left\mid 4 - {x}^{2} \right\mid + C$