According to the Wolfram integrator: $$\int (x^2+a)^{-3/2}\text{d}x = {x \over a \times \sqrt{(x^2+a)}}$$
I easily differentiated the answer to verify that it was correct (not that I don't trust Wolfram or anything), but how would I get this result on my own? I'm reasonably sure that integration by parts should be used, although I tried splitting it up a few different ways and I couldn't get it to work. A short hint would probably suffice. (You might be able to guess what physics problem I was working on when I encountered this integral).
HINT:
Assuming all variables are postitive:
$$\int\left(x^2+a\right)^{-\frac{3}{2}}\space\text{d}x=$$ $$\int\frac{1}{\left(x^2+a\right)^{\frac{3}{2}}}\space\text{d}x=$$
Substitute $x=\sqrt{a}$ and $\text{d}x=\sqrt{a}\sec^2(u)\space\text{d}u$. Then $\left(x^2+a\right)^{\frac{3}{2}}=\left(a\tan^2(u)+a\right)^{\frac{3}{2}}=a^{\frac{3}{2}}\sec^3(u)$ and $u=\arctan\left(\frac{x}{\sqrt{a}}\right)$:
$$\sqrt{a}\int\frac{\cos(u)}{a^{\frac{3}{2}}}\space\text{d}u=$$ $$\frac{1}{a}\int\cos(u)\space\text{d}u$$