Answer
$$\ln \left| {6 + 3\tan t} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{3{{\sec }^2}t}}{{6 + 3\tan t}}} dt \cr
& {\text{Use substitution:}}\cr
& {\text{Let }}u = 6 + 3\tan t,{\text{ so that }}du = 3{\sec ^2}tdt \cr
& dt = \frac{{du}}{{3{{\sec }^2}t}} \cr
& {\text{write the integral in terms of }}u \cr
& \int {\frac{{3{{\sec }^2}t}}{{6 + 3\tan t}}} dt = \int {\frac{{3{{\sec }^2}t}}{u}} \left( {\frac{{du}}{{3{{\sec }^2}t}}} \right) \cr
& = \int {\frac{1}{u}} \left( {\frac{{du}}{1}} \right) \cr
& = \int {\frac{1}{u}} du \cr
& {\text{integrate }} \cr
& = \ln \left| u \right| + C \cr
& {\text{write in terms of }}t{\text{; replace }}6 + 3\tan t{\text{ for }}u \cr
& = \ln \left| {6 + 3\tan t} \right| + C \cr} $$
