Simple Integral. I'm a bit rusty.

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Homework Statement



Take the integral of
1/[(z^2 + x^2) ^ 3/2] dx where z is a constant.



Homework Equations





The Attempt at a Solution



I tried substitution, but couldn't get it to work. Thanks in advance.
 
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Have you tried x=ztanθ ?
 
Yes but I was confused on the next step. I tried to look into trig substitutions.
 
Could you post your entire working?
 
Oh, never mind I think I just figured it out. Thanks for your help!
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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