Answer
$$y = \tan \left( {\frac{{{x^3}}}{3} + C} \right)$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \left( {1 + {y^2}} \right){x^2} \cr
& {\text{separating the variables}} \cr
& \frac{{dy}}{{1 + {y^2}}} = {x^2}dx \cr
& \cr
& {\text{integrate both sides }} \cr
& \int {\frac{{dy}}{{1 + {y^2}}}} = \int {{x^2}} dx \cr
& {\tan ^{ - 1}}y = \frac{{{x^3}}}{3} + C \cr
& {\text{solving for }}y \cr
& \tan \left( {{{\tan }^{ - 1}}y} \right) = \tan \left( {\frac{{{x^3}}}{3} + C} \right) \cr
& y = \tan \left( {\frac{{{x^3}}}{3} + C} \right) \cr} $$
