Answer
$$\ln \left| {2 + \sec y} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sec y\tan y}}{{2 + \sec y}}} dy \cr
& {\text{Use substitution:}}\cr
& {\text{Let }}u = 2 + \sec y,{\text{ so that }}du = \sec y\tan ydy \cr
& {\text{Write the integral in terms of }}u \cr
& \int {\frac{{\sec y\tan y}}{{2 + \sec y}}} dy = \int {\frac{{\sec y\tan y}}{u}} \left( {\frac{{dy}}{{\sec y\tan y}}} \right) \cr
& = \int {\frac{1}{u}} \left( {\frac{{du}}{1}} \right) \cr
& = \int {\frac{1}{u}} du \cr
& {\text{Integrate }} \cr
& = \ln \left| u \right| + C \cr
& {\text{Write in terms of }}y:\cr
& {\text{Replace }}2 + \sec y\tan t{\text{ for }}u \cr
& = \ln \left| {2 + \sec y} \right| + C \cr} $$
