Originally Posted by

**M1ke** I am trying to integrate the following problem with limit +D and -D

$\displaystyle \frac{1}{(z^2 + r^2)^\frac{3}{2}}$

Now intuition tells me to treat it like

$\displaystyle (z^2 + r^2)^\frac{-3}{2}$

Here is where I'm going wrong - the way I'm integrating gives me

$\displaystyle \frac{(2z)(z^2 + r^2)^\frac{-1}{2}}{\frac{-1}{2}}$

$\displaystyle \frac{-4z}{(z^2 + r^2)^\frac{-1}{2}}$

But if I differentiate that I get

$\displaystyle \frac{4r^2}{(z^2 + r^2)^\frac{3}{2}}$

So have randomly created a factor of $\displaystyle 4r^2$. Therefore I'm doing something wrong but I can't seem to get a rule to fit the situation.

Differentiating

$\displaystyle \frac{z}{r^2(z^2 + r^2)^\frac{1}{2}}$

Returns me to the original problem

$\displaystyle \frac{1}{(z^2 + r^2)^\frac{3}{2}}$

But that would imply that the basic integral is divided by $\displaystyle r^2$ despite the fact that for all intents and purposes $\displaystyle r^2$ is a constant.

Can anyone explain, as I think I've horribly confused myself about something not that complex?