## Calculus 10th Edition

$-\frac{1}{3(1+x^{3})}+C$
$let$ $1+x^3=u$ $du=3x^2$ $dx$ $\int$$u^{-2} du =\frac{1}{3}\int$$(1+x^3)^{-2}$ $(3x^2)$ $dx$ $=\frac{1}{3}(\frac{(1+x^{3})^{-1}}{-1})+C$ $=-\frac{1}{3(1+x^{3})}+C$