Originally Posted by

**skeeter** $\displaystyle \int \frac{x^3}{\sqrt{x^2+1}} \, dx$

integration by parts ...

$\displaystyle u = x^2$

$\displaystyle dv = \frac{x}{\sqrt{x^2+1}} \, dx

$

$\displaystyle du = 2x \, dx$

$\displaystyle v = \sqrt{x^2+1}$

$\displaystyle x^2 \sqrt{x^2+1} - \int 2x\sqrt{x^2+1} \, dx

$

$\displaystyle x^2\sqrt{x^2+1} - \frac{2}{3}\sqrt{(x^2+1)^3} + C

$

$\displaystyle \sqrt{x^2+1}\left[x^2 - \frac{2}{3}(x^2+1)\right] + C$

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