# What is the integral of x^3/(x^2+1)?

Oct 9, 2016

$\int {x}^{3} / \left({x}^{2} + 1\right) \mathrm{dx} = \frac{{x}^{2} - \ln \left({x}^{2} + 1\right)}{2} + C$

#### Explanation:

We will use integration by substitution, as well as the integrals $\int \frac{1}{x} \mathrm{dx} = \ln | x | + C$ and $\int 1 \mathrm{dx} = x + C$

$\int {x}^{3} / \left({x}^{2} + 1\right) \mathrm{dx} = \int {x}^{2} / \left({x}^{2} + 1\right) x \mathrm{dx}$

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

Let $u = {x}^{2} + 1 \implies \mathrm{du} = 2 x \mathrm{dx}$. Then

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

$= \frac{1}{2} \int \left(1 - \frac{1}{u}\right) \mathrm{du}$

$= \frac{1}{2} \left(u - \ln | u |\right) + C$

$= \frac{{x}^{2} + 1}{2} - \ln \frac{{x}^{2} + 1}{2} + C$

$= {x}^{2} / 2 - \ln \frac{{x}^{2} + 1}{2} + \frac{1}{2} + C$

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

(Note that as $C$ is an arbitrary constant, we can disregard the $\frac{1}{2}$ as we did in the last step. Adding an additional constant makes no difference when we already are considering all functions of that form which vary by a constant.)