Answer
$$\frac{1}{3}\ln \left| {1 + {x^3}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{1 + {x^3}}}} dx \cr
& {\text{Integrate by using the substitution method}} \cr
& \,\,\,{\text{Let }}u = 1 + {x^3},\,\,\,\,du = 3{x^2}dx,\,\,\,\,dx = \frac{{du}}{{3{x^2}}} \cr
& \cr
& {\text{Write the integrand in terms of }}u \cr
& \int {\frac{{{x^2}}}{{1 + {x^3}}}} dx = \int {\frac{{{x^2}}}{u}} \left( {\frac{{du}}{{3{x^2}}}} \right) \cr
& = \frac{1}{3}\int {\frac{1}{u}} du \cr
& \cr
& {\text{Integrating}} \cr
& = \frac{1}{3}\ln \left| u \right| + C \cr
& {\text{Write in terms of }}x;{\text{ substitute }}1 + {x^3}{\text{ for }}u \cr
& = \frac{1}{3}\ln \left| {1 + {x^3}} \right| + C \cr} $$
