Question:
$\frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}}$
Solution:
Let $\tan x=t$
$\therefore \sec ^{2} x d x=d t$
$\Rightarrow \int \frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}} d x=\int \frac{d t}{\sqrt{t^{2}+2^{2}}}$
$=\log \left|t+\sqrt{t^{2}+4}\right|+C$
$=\log \left|\tan x+\sqrt{\tan ^{2} x+4}\right|+\mathrm{C}$