Prove

$\frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}$

$=\frac{\cos 2 x+(1-\cos 2 x)}{\cos ^{2} x}$     $\left[\cos 2 x=1-2 \sin ^{2} x\right]$

$=\frac{1}{\cos ^{2} x}$

$=\sec ^{2} x$

$\therefore \int \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} d x=\int \sec ^{2} x d x=\tan x+\mathrm{C}$