Answer
$$y = C{e^{\frac{{{x^3}}}{3} + x}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = y\left( {{x^2} + 1} \right) \cr
& {\text{separate}} \cr
& \frac{{dy}}{y} = \left( {{x^2} + 1} \right)dx \cr
& {\text{integrate both sides}} \cr
& \int {\frac{{dy}}{y}} = \int {\left( {{x^2} + 1} \right)} dx \cr
& \ln \left| y \right| = \frac{{{x^3}}}{3} + x + c \cr
& {\text{solve for }}y \cr
& {e^{\ln \left| y \right|}} = {e^{\frac{{{x^3}}}{3} + x + c}} \cr
& y = {e^c}{e^{\frac{{{x^3}}}{3} + x}} \cr
& {\text{set }}C = {e^c} \cr
& y = C{e^{\frac{{{x^3}}}{3} + x}} \cr} $$
