Answer
$$ - {e^{ - {t^2}}} + C $$
Work Step by Step
$$\eqalign{
& \int {2t{e^{ - {t^2}}}} dt \cr
& {\text{set }}u = - {t^2}{\text{ then }}\frac{{du}}{{dt}} = - 2t \to \frac{{du}}{{ - 2t}} = dt \cr
& {\text{write the integrand in terms of }}u \cr
& \int {2t{e^{ - {t^2}}}} dt = \int {2t{e^u}} \left( {\frac{{du}}{{ - 2t}}} \right) \cr
& {\text{cancel common terms}} \cr
& = - \int {{e^u}} du \cr
& {\text{integrating}} \cr
& = - {e^u} + C \cr
& {\text{replace }} - {t^2}{\text{ for }}u \cr
& = - {e^{ - {t^2}}} + C \cr} $$
