## Calculus: Early Transcendentals (2nd Edition)

$\frac{1}{3}\ln (y^3+27)+c$
$\int\frac{y^2}{y^3+27}dy$ Use u-substitution... $u=y^3+27$ $dy=\frac{du}{3y^2}$ $\int\frac{y^2}{u}\frac{du}{3y^2}=\frac{1}{3}\int\frac{1}{u}du$ $=\frac{1}{3}\ln u+c$ substitute out u... $\frac{1}{3}\ln (y^3+27)+c$