Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 5

$\int (3x+2)(3x^2+4x)^4 dx=\frac{(3x^2+4x)^5}{10}+C$

$\int (3x+2)(3x^2+4x)^4 dx$ $\space$ and $\space \space u = 3x^2 + 4x$ $du = (6x +4)dx$ Doing the substitution $\space\space u = 3x^2 + 4x$ $\int (u)^4 \frac{du}{2}\space$ => $\space\frac{1}{2}\int u^4 du$ $\frac{1}{2}\int u^4 du = \frac{u^5}{10} + C$ Backing to x $\int (3x+2)(3x^2+4x)^4 dx=\frac{(3x^2+4x)^5}{10}+C$