I'm stuck integrating this:

$\displaystyle \displaystyle\int\frac{1}{(x^2+3)^{\frac{3}{2}}} \ dx$

I really just need a hint on how to start this question.

EDIT: I think I worked out how to do it

Let $\displaystyle x=\sqrt{3}\tan\theta \Rightarrow dx=\sqrt{3}\sec^2\theta \ d\theta$

$\displaystyle =\displaystyle\int\frac{\sqrt{3}\sec^2\theta}{(3\t an^2\theta+3)^{\frac{3}{2}}} \ d\theta$

$\displaystyle =\displaystyle\int\frac{\sqrt{3}\sec^2\theta}{3^{\ frac{3}{2}}\sec^3\theta} \ d\theta$

$\displaystyle =\displaystyle\int\frac{1}{3\sec\theta} \ d\theta$

$\displaystyle =\displaystyle\frac{1}{3}\int\cos\theta \ d\theta$

$\displaystyle =\displaystyle\frac{1}{3}\sin\theta+C$

$\displaystyle \dfrac{x}{\sqrt{3}}=\tan\theta \Rightarrow \sin\theta=\dfrac{x}{\sqrt{x^2+3}}$

$\displaystyle =\dfrac{x}{3\sqrt{x^2+3}}+C$