Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 275: 30

$$\frac{1}{15}(9t+2)^{5/3}+c$$

Given $$ \int(9 t+2)^{2 / 3} d t$$ Let $$u=9t+2\ \ \ \Rightarrow \ \ du=9dt $$ then \begin{align*} \int(9 t+2)^{2 / 3} d t&=\frac{1}{9}\int(u)^{2 / 3} d t\\ &=\frac{1}{9}\frac{3}{5}u^{5/3}+c\\ &=\frac{1}{15}(9t+2)^{5/3}+c \end{align*}