Originally Posted by

**galactus** You could use trig sub.

$\displaystyle \int x^{3}\sqrt{x^{2}+1}dx$

Let $\displaystyle x=tan(t), \;\ dx=sec^{2}(t)dt$

Upon making the subs, we get:

$\displaystyle \int tan^{3}(t)sec^{3}(t)dt$

$\displaystyle \int tan^{2}(t)sec^{2}(t)(sec(t)tan(t)dt$

=$\displaystyle \int (sec^{2}(t)-1)sec^{2}(t)(sec(t)tan(t))dt$

Let $\displaystyle u=sec(t), \;\ du=sec(t)tan(t)dt$

$\displaystyle \int (u^{2}-1)u^{2}du$

$\displaystyle \frac{1}{5}u^{5}-\frac{1}{3}u^{3}$

Resub:

$\displaystyle \frac{1}{5}sec^{5}(t)-\frac{1}{3}sec^{3}(t)$

Resub from the beginning, $\displaystyle t=tan^{-1}(x)$

This gives:

$\displaystyle \frac{1}{5}(x^{2}+1)^{\frac{5}{2}}-\frac{1}{3}(x^{2}+1)^{\frac{3}{2}}+C$

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An easier way may be to let $\displaystyle u=x^{2}+1, \;\ du=2xdx, \;\ \frac{1}{2}du=dx$

Then we get:

$\displaystyle \frac{1}{2}\int \sqrt{u}(u-1)du=\frac{1}{2}\left[\int u^{\frac{3}{2}}-\int u^{\frac{1}{2}}\right]du$

Can you go from there?. This way is easier if the trig sub is scary.