Integrate Form $du / (a^2 + u^2)^{3/2}$

A trigonometric substitution does indeed work.

We want to express $(a^2 + u^2)^{3/2}$ as something without square roots. We want to use some form of the Pythagorean trigonometric identity $\sin^2 x + \cos^2 x = 1$. Multiplying each side by $\frac{a^2}{\cos^2 x}$, we get $a^2 \tan^2 x + a^2 = a^2 \sec^2 x$, which is in the desired form. of (sum of two squares) = (something squared).

This suggests that we should use the substitution $u^2 = a^2 \tan^2 x$. Equivalently, we substitute $u = a \tan x$ and $du = a \sec^2 x dx$. Then $$ \int \frac{du}{(a^2 + u^2)^{3/2}} = \int \frac{a \sec^2 x \, dx}{(a^2 + a^2 \tan^2 x)^{3/2}}. $$ Applying the trigonometric identity considered above, this becomes $$ \int \frac{a \sec^2 x \, dx}{(a^2 \sec^2 x)^{3/2}} = \int \frac{dx}{a^2 \sec x} = \frac{1}{a^2} \int \cos x \, dx, $$ which can be easily integrated as $$ =\frac{1}{a^2} \sin x. $$ Since we set $u = a \tan x$, we substitute back $x = \tan^{-1} (\frac ua)$ to get that the answer is $$ =\frac{1}{a^2} \sin \tan^{-1} \frac{u}{a}. $$ Since $\sin \tan^{-1} z = \frac{z}{\sqrt{z^2 + 1}}$, this yields the desired result of $$ =\frac{u/a}{a^2 \sqrt{(u/a)^2 + 1}} = \frac{u}{a^2 (a^2 + u^2)^{1/2}}. $$

Let $u = a \tan \theta$ and work from there. See this (page about trig substitutions).