Help with this integral? I can't figure out the substitution. I've been struggling to figure out this integral. 
$$\int \frac {1}{x\sqrt{5-x^2}}$$
I'm almost certain it has something to do with this fact:
$$\int \frac 1{\sqrt{1-x^2}} = \sin^{-1}(x) + C$$
But I can't figure out how to use that to my advantage. No obvious substitution is jumping out at me either. I'm guessing there's something simple I'm missing... any help will be much appreciated!
Edit (11/13/13): An update on this, just for the sake of finishing what I started - I had gotten a bit stuck on this problem and could not seem to figure it out until I saw it worked out in this thread on reddit. Where I was stuck turned out to be a very simple algebra manipulation. Just thought I'd add that link for the sake of any Googlers or what have you who might find this post and want to see it worked out in more detail. Many thanks to all involved in helping me understand this problem! 
 A: A natural substitution is $u=\sqrt{5-x^2}$, though I prefer the version $u^2=5-x^2$. Then $2u\,du=-2x\,dx$.
Rewrite our integral as 
$$\int \frac{x\,dx}{x^2\sqrt{5-x^2}}.$$
After we make the substitution, we end up with
$$\int \frac{du}{u^2-5}.$$
Factor $u^2-5$ as $(u-\sqrt{5})(u+\sqrt{5})$ and use partial fractions. 
A: So substitute $ \sqrt{5-x^2} $ as u therefore ${u^2} = 5-x^2 $ right. From this keep in mind that x can be made subject of the formula and that will make $ x= \sqrt{5-u^2} $ .
$ -2x dx = 2u du$ therefore $ x dx = - u du$ . You can multiply the equation by 1,which in fundamentals of mathematics there is no change made. 1 can be written as $ \frac{x}{x} $ making the equation $ -1 \int \frac{x u du}{x^2 u} $. After cancelling out the u s go and one x remains but it is with respect to u (the du part) so use the $x=\sqrt{5-u^2} $ for x. The x can be further broken down into $ \sqrt{\sqrt 5^2}-u^2 $ and then use the arcsin formula which is $\int \frac{1}{\sqrt{a^2-x^2}}$ = $arcsin \frac{x}{a} + C$,voila.
