Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 5

$\ln |6+3 \tan t|+c$

Solve $\int\dfrac{3 \sec^2 t}{6+3 \tan t}$ Let $p=6+3 \tan t$ and $dp=3 \sec^2 t$ This implies $\int\dfrac{3 \sec^2 t}{6+3 \tan t}=\int\dfrac{dp}{p}=\ln|p|+c$ Since, $p=6+3 \tan t$ Thus, $\int\dfrac{3 \sec^2 t}{6+3 \tan t}=\ln |6+3 \tan t|+c$