Simplifying Integrands: How to Factor Out Constants in the Denominator?

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This equality is given in an example, in my textbook:

\int\frac{1}{a^2+x^2}dx=\frac{1}{a^2}\int\frac{1}{1+(\frac{x}{a})^2}dx

So, my question is simply... how?
This is really more of an algebra question, but it'd really help me to see some more detail here.
 
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\frac{1}{a^2+x^2} = \frac{1}{a^2(1+ \frac{x^2}{a^2})} = \frac{1}{a^2}\frac{1}{1+ (\frac{x}{a})^2}
 
I see that... sort of. It's a form of factoring, obviously.
I don't quite get the \frac{x^2}{a^2} bit, though...
Well, wait: a^2\times1=a^2 obviously, and a^2\times\frac{x^2}{a^2}=x^2 right? Neat trick, there. Completely obvious too, now that I think about it, but... I guess that I just didn't realize that could be done. Weird.
 
Just think of it as factoring out an a2 from the bottom. Here's an easier example:
\frac{1}{4+x^{2}}=\frac{1}{(4)1+\frac{x^{2}}{4}}=(\frac{1}{4})\frac{1}{1+{\frac{x^{2}}{4}}}
 
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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