## Thomas' Calculus 13th Edition

$\frac{1}{3}tan(3x+2)+C$
Let u=3x+2 $=>du=3dx=>\frac{1}{3}du=dx$ =$\int sec^2(3x+2)dx$ =$\int (sec^2u)(\frac{1}{3}du)$ =$\frac{1}{3}\int sec^2udu$ =$\frac{1}{3}tan u+C$ =$\frac{1}{3}tan(3x+2)+C$