Answer
$$ - \frac{1}{2}{e^{ - {r^2}}} + C$$
Work Step by Step
$$\eqalign{
& \int {r{e^{ - {r^2}}}} dr \cr
& {\text{set }}u = - {r^2}{\text{ then }}\frac{{du}}{{dr}} = - 2r \to rdr = - \frac{1}{2}du \cr
& {\text{write the integrand in terms of }}u \cr
& \int {r{e^{ - {r^2}}}} dr = \int {{e^u}} \left( { - \frac{1}{2}dr} \right) \cr
& = - \frac{1}{2}\int {{e^u}} dr \cr
& {\text{integrating}} \cr
& = - \frac{1}{2}{e^u} + C \cr
& {\text{replace }} - {r^2}{\text{ for }}u \cr
& = - \frac{1}{2}{e^{ - {r^2}}} + C \cr} $$
