Answer
$${e^{{x^2}}} + C$$
Work Step by Step
$$\eqalign{
& \int {2x{e^{{x^2}}}} dx \cr
& {\text{set }}u = {x^2}{\text{ then }}\frac{{du}}{{dx}} = 2x,\,\,\,\,\,\,du = 2xdx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {2x{e^{{x^2}}}} dx = \int {{e^u}} du \cr
& {\text{integrating by the formula }}\int {{e^u}} du = {e^u} + C \cr
& = {e^u} + C \cr
& {\text{replace }}{x^2}{\text{ for }}u \cr
& = {e^{{x^2}}} + C \cr} $$
