Answer
$$y = \tan \left( {x + \frac{{{x^2}}}{2} + C} \right)$$
Work Step by Step
$$\eqalign{
& y' - \left( {1 + x} \right)\left( {1 + {y^2}} \right) = 0 \cr
& {\text{write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} - \left( {1 + x} \right)\left( {1 + {y^2}} \right) = 0 \cr
& \cr
& {\text{separating the variables}} \cr
& \frac{{dy}}{{dx}} = \left( {1 + x} \right)\left( {1 + {y^2}} \right) \cr
& \frac{{dy}}{{1 + {y^2}}} = \left( {1 + x} \right)dx \cr
& \cr
& {\text{integrate both sides }} \cr
& \int {\frac{{dy}}{{1 + {y^2}}}} = \int {\left( {1 + x} \right)} dx \cr
& {\tan ^{ - 1}}y = x + \frac{{{x^2}}}{2} + C \cr
& \cr
& {\text{solving for }}y \cr
& y = \tan \left( {x + \frac{{{x^2}}}{2} + C} \right) \cr} $$
