Difficult Integral: $\int\frac{x^n}{\sqrt{1+x^2}}dx$ How to calculate this difficult integral: $\int\frac{x^2}{\sqrt{1+x^2}}dx$?
The answer is $\frac{x}{2}\sqrt{x^2\pm{a^2}}\mp\frac{a^2}{2}\log(x+\sqrt{x^2\pm{a^2}})$.
And how about $\int\frac{x^3}{\sqrt{1+x^2}}dx$?

Recall the hyperbolic functions 
$$\cosh t= \frac{e^t + e^{-t}}{2} = \cos(it)$$ 
and $$\sinh t=\frac{e^t - e^{-t}}{2} = i\sin(-it).$$
Note that $\frac{d}{dt}\sinh t = \cosh t$, $\frac{d}{dt}\cosh t = \sinh t$ and also $\cosh^2 t -\sinh^2 t = 1$.
Making the substitution $\sinh t=x $ we see that
$$\frac{x^n\, dx}{\sqrt{1+x^2}} = \frac{\sinh^n t\, \cosh t\,dt}{\sqrt{1+\sinh^2t}}=
\frac{\sinh^n t\, \cosh t\,dt}{\sqrt{\cosh^2t}}=\sinh^n t\, dt$$
which leads us to $$\int\frac{x^n\, dx}{\sqrt{1+x^2}} = \int \sinh^n t\, dt.$$
To complete the problem, the binomial theorem is useful.
