Find the integral of the function $\frac{\cos 2x}{(\cos x + \sin x)^2}$.

(D) We have the integral $I = \int \frac{\cos 2x}{(\cos x + \sin x)^2} dx$.
Using the trigonometric identity $\cos 2x = \cos^2 x - \sin^2 x$ and the expansion $(\cos x + \sin x)^2 = \cos^2 x + \sin^2 x + 2 \sin x \cos x = 1 + \sin 2x$,we get:
$I = \int \frac{\cos^2 x - \sin^2 x}{1 + \sin 2x} dx = \int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2} dx$
$I = \int \frac{\cos x - \sin x}{\cos x + \sin x} dx$
Let $u = \cos x + \sin x$. Then $du = (-\sin x + \cos x) dx$.
Substituting these into the integral:
$I = \int \frac{1}{u} du = \log |u| + C$
$I = \log |\cos x + \sin x| + C$,where $C$ is an arbitrary constant.