## University Calculus: Early Transcendentals (3rd Edition)

$\frac{(3x^2+4x)^5}{10}+c$
$u=3x^2+4x$, therefore $du=(6x+4)dx=2(3x+2)dx$. We do the substitution! $\int((3x+2)(3x^2+4x)^4) dx=\int u^4\frac{du}{2}=\frac{1}{2}\int u^4du=\frac{1}{2}\frac{u^5}{5}+c=\frac{u^5}{10}+c$ We re-do the substitution! $\frac{u^5}{10}+c=\frac{(3x^2+4x)^5}{10}+c$