Answer
$y = \tan \left( {{x^2} + C} \right)$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = 2x\left( {{y^2} + 1} \right) \cr
& {\text{Separating the variables}} \cr
& \frac{{dy}}{{{y^2} + 1}} = 2xdx \cr
& {\text{Integrate both sides of the equation}} \cr
& \int {\frac{{dy}}{{{y^2} + 1}}} = \int {2x} dx \cr
& \arctan y = {x^2} + C \cr
& {\text{solve for }}y \cr
& \tan \left( {\arctan y} \right) = \tan \left( {{x^2} + C} \right) \cr
& y = \tan \left( {{x^2} + C} \right) \cr} $$
