Answer
$$\ln \left| {\sec x + 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin x}}{{\cos x + {{\cos }^2}x}}} dx \cr
& {\text{Let }}t = \cos x,{\text{ }}dt = - \sin xdx \cr
& {\text{Substituting}} \cr
& \int {\frac{{\sin x}}{{\cos x + {{\cos }^2}x}}} dx = - \int {\frac{1}{{t + {t^2}}}} du \cr
& {\text{Decompose into partial fractions}} \cr
& \frac{1}{{t + {t^2}}} = \frac{1}{{t\left( {1 + t} \right)}} = \frac{A}{t} + \frac{B}{{1 + t}} \cr
& 1 = A\left( {1 + t} \right) + Bt \cr
& t = 0 \to A = 1 \cr
& t = - 1 \to B = - 1 \cr
& \frac{1}{{t\left( {1 + t} \right)}} = \frac{1}{t} - \frac{1}{{1 + t}} \cr
& - \int {\frac{1}{{t + {t^2}}}} du = \int {\left( {\frac{1}{{1 + t}} - \frac{1}{t}} \right)} du \cr
& {\text{Integrate}} \cr
& {\text{ = ln}}\left| {1 + t} \right| - \ln \left| t \right| + C \cr
& {\text{Write in terms of }}x \cr
& {\text{ = ln}}\left| {1 + \cos x} \right| - \ln \left| {\cos x} \right| + C \cr
& = \ln \left| {\frac{{1 + \cos x}}{{\cos x}}} \right| + C \cr
& = \ln \left| {\sec x + 1} \right| + C \cr} $$
