Answer
$\frac{cot^2\ u}{8sec\ u}$
Work Step by Step
$\frac{1}{x^2\sqrt(4 + x^2)}$ , $x = 2 tan\ u$
We know that $1+tan^2\ u=sec^2\ u$
$\frac{1}{x^2\sqrt(4 + x^2)}=\frac{1}{((2 tan\ u)^2\sqrt(4 + (2 tan\ u)^2)}=\frac{1}{(4tan^2\ u)\sqrt(4 + 4tan^2\ u)}=\frac{1}{(8tan^2\ u)\sqrt(1 + tan^2\ u)}=\frac{1}{(8tan^2\ u)\sqrt(sec^2\ u)}\frac{1}{8tan^2\ u\ sec\ u}=\frac{cot^2\ u}{8sec\ u}$
