# How do you find the antiderivative of int x^3/(4+x^2) dx?

Dec 2, 2016

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

Let $u = 4 + {x}^{2}$, implying that $\mathrm{du} = 2 x \mathrm{dx}$. Also note that ${x}^{2} = u - 4$. Rearranging the integral:

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

Splitting up the integral:

$I = \frac{1}{2} \int \mathrm{du} - 2 \int \frac{1}{u} \mathrm{du} = \frac{1}{2} u - 2 \ln \left\mid u \right\mid = \frac{1}{2} \left(4 + {x}^{2}\right) - 2 \ln \left\mid 4 + {x}^{2} \right\mid + C$

Letting the constant from $\frac{1}{2} \left(4 + {x}^{2}\right)$ absorb into $C$:

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