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

Here's one way to go about deriving a recursion relation for integrals of the form
$$\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx$$
Split the integral like so:
$$\int x^{n-1}\frac{x}{\sqrt{1+x^2}}\mathrm dx$$
and integrate by parts:
$$\int x^{n-1}\frac{x}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\int\sqrt{1+x^2} x^{n-2}\mathrm dx$$
Noting that $1+x^2$ is always positive for real $x$, we then complicate things a little:
$$\int \frac{x^n}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\int(1+x^2)\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx$$
Perform another split:
$$\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\left(\int \frac{x^n}{\sqrt{1+x^2}}\mathrm dx+\int\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx\right)$$
and we see something we can isolate:
$$n\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx=x^{n-1}\sqrt{1+x^2}-(n-1)\int\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx$$
and then we finally divide both sides by $n$:
$$\int\frac{x^n}{\sqrt{1+x^2}}\mathrm dx=\frac1{n}\left(x^{n-1}\sqrt{1+x^2}-(n-1)\int\frac{x^{n-2}}{\sqrt{1+x^2}}\mathrm dx\right)$$
We can use the starting values $\int\frac{\mathrm dx}{\sqrt{1+x^2}}=\mathrm{arsinh}\,x$ and $\int\frac{x \mathrm dx}{\sqrt{1+x^2}}=\sqrt{1+x^2}$ for the recursion.
(This is a response to Srivatsan's comment, which got too long for the comment box.)
