$\int \sqrt { \sin ^2 2x +\cos ^2 2x + 2 \sin 2x \cos 2 x }dx$
$\sin ^2 \alpha + \cos ^2 \alpha=1$
$\sin 2 \alpha = 2 \sin \alpha \cos \alpha$
$\int \sqrt { (\sin 2x+ \cos 2x)^2}$$dx$
$\int (\sin 2x+ \cos 2x) dx$
$ \large\frac{-\cos 2x}{2}+\frac{\sin 2x}{2}$$+c $
Hence d is the correct answer.