Solving Bizarre Integral: x/[z2(x2+z2)1/2]

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Now then, I am close to shedding a tear with this one.

This integral has been popping up in a few electromag examples iv been doing and i have absolutely no idea what's going on here.

The integral is 1/[(x2+z2)3/2] with respect to x

According to the textbook the answer is x/[z2(x2+z2)1/2]

I initially, without evening really thinking, went straight for -1/(x[x2+z2]1/2)

Any ideas?
Thanks
 
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Use the substitution x=z \tan u.
 
btw, you left a constant
 

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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