Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 23

$$\int\sec^2(3x+2)dx=\frac{1}{3}\tan(3x+2)+C$$

$$A=\int\sec^2(3x+2)dx$$ We set $u=3x+2$. Then $$du=3dx$$ That means, $$dx=\frac{1}{3}du$$ Therefore, $$A=\frac{1}{3}\int\sec^2udu=\frac{1}{3}\tan u+C$$ $$A=\frac{1}{3}\tan(3x+2)+C$$