Variable change in definite integration

bartrocs
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I am having trouble understanding how the author of the textbook "Physics for scientists and engineers" (Randall D Knight), is performing the variable change in the definite integral of the picture that I have taken a screen clipping of. Can someone please help me figure this out?
 

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Variable r is being replaced by variable u where

u = z^{2} + r^{2}.

Hence

\frac{du}{dr} = 2r

since z is a constant.
 
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thanks! I understand now
 
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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