Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 276: 55

$$\frac{1}{4} \tan (4 x+9)+C$$

Given $$ \int \sec ^{2}(4 x+9) d x$$ Let $$u=4 x+9 \ \ \ \ \Rightarrow \ \ \ du =4dx $$ \begin{aligned} \int \sec ^{2}(4 x+9) d x &=\frac{1}{4} \int \sec ^{2} u d u \\ &=\frac{1}{4} \tan u+C \\ &=\frac{1}{4} \tan (4 x+9)+C \end{aligned}