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

Jun 24, 2016

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

Explanation:

We have the integral:

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

We will use substitution: let $u = {x}^{2} + 1$, implying that $\mathrm{du} = 2 x \mathrm{dx}$.

Rearrange the integral, including making $2 x \mathrm{dx}$ present:

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

Make the following substitutions into the integral:

$\left\{\begin{matrix}{x}^{2} + 1 = u \\ {x}^{2} = u - 1 \\ 2 x \mathrm{dx} = \mathrm{du}\end{matrix}\right.$

We obtain:

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

Splitting the integral up through subtraction:

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

These are common integrals:

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

Since $u = {x}^{2} + 1$:

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

The absolute value bars are not needed since ${x}^{2} + 1 > 0$ for all Real values of $x$. Also note that $\frac{1}{2} \left({x}^{2} + 1\right) = \frac{1}{2} {x}^{2} + \frac{1}{2}$, so the $\frac{1}{2}$ gets absorbed into the constant of integration $C$:

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

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