How do I calculate an integral with a square root in the denominator?

[solkahns]

Hi,
I have a problem. I can't understand how it becomes [-1/the square root of (z^2+r^2)] from integral (r.dr/(z^2+r^2)^3/2)
It seems complicated but I couldn't write it different.
Thanks, Sol
P.S. The integral goes to R from 0.

 

[quantumdude]

It's a simple u-substitution. Try to do it, and come back if you get stuck.

 

[arildno]

$$\frac{d}{dr}(-\frac{1}{(z^{2}+r^{2})^{\frac{1}{2}}})=\frac{1}{2}*\frac{1}{(z^{2}+r^{2})^{\frac{3}{2}}}*2r=\frac{r}{(z^{2}+r^{2})^{\frac{3}{2}}}$$
That's all there is to it, really..