Question

Evaluate:
$\int \frac{x^3}{\sqrt{1+x^2}}dx$

Answer 1

Let $I=\int \frac{x^3}{\sqrt{1+x^2}}dx$.

$I=\int \frac{x^2\cdot x}{\sqrt{1+x^2}}dx$

Put $1+x^2=u$, then,

$x^2=u-1$

$2xdx=du$

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

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

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

$=\frac{1}{2}\left(\frac{2}{3}{u}^{3/2}\right)-\frac{1}{2}\left(2u^{1/2}\right)+c$

$=\frac{1}{3}{u}^{3/2}-u^{1/2}+c$

$=u^{1/2}\left(\frac{1}{3}u-1\right)+c$

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

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

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

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