Answer
$\frac{1}{3}\ln (y^3+27)+c$
Work Step by Step
$\int\frac{y^2}{y^3+27}dy$
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...
$\frac{1}{3}\ln (y^3+27)+c$
