Answer
$\frac{1}{2}{\tan ^2}2x - \frac{1}{2}\sec 2x + C$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{{\left( {\sin x + \cos x} \right)}^2}}}} dx \cr
& {\text{Expanding the binomial}} \cr
& = \int {\frac{1}{{{{\sin }^2}x + 2\sin x\cos x + {{\cos }^2}x}}} dx \cr
& = \int {\frac{1}{{1 + 2\sin x\cos x}}} dx \cr
& {\text{Recall that }}\sin 2x = 2\sin x\cos x \cr
& = \int {\frac{1}{{1 + \sin 2x}}} dx \cr
& = \int {\frac{1}{{1 + \sin 2x}} \times \left( {\frac{{1 - \sin 2x}}{{1 - \sin 2x}}} \right)} dx \cr
& {\text{Rationalizing}} \cr
& = \int {\frac{{1 - \sin 2x}}{{1 - {{\sin }^2}2x}}} dx \cr
& {\text{Using the pythagorean identity co}}{{\text{s}}^2}\theta = 1 - {\sin ^2}\theta \cr
& = \int {\frac{{1 - \sin 2x}}{{{{\cos }^2}2x}}} dx \cr
& = \int {\left( {\frac{1}{{{{\cos }^2}2x}} - \frac{{\sin 2x}}{{{{\cos }^2}2x}}} \right)} dx \cr
& = \int {\left( {{{\sec }^2}2x - \sec 2x\tan 2x} \right)} dx \cr
& {\text{Integrating}} \cr
& = \frac{1}{2}{\tan }2x - \frac{1}{2}\sec 2x + C \cr} $$