Answer
$$ - \frac{1}{2}{e^{1 - {x^2}}} + C$$
Work Step by Step
$$\eqalign{
& \int {x{e^{1 - {x^2}}}} dx \cr
& {\text{Integrate by using the substitution method}} \cr
& {\text{Let: }}u = 1 - {x^2},{\text{ }}du = - 2xdx,{\text{ }}dx = - \frac{1}{{2x}}du \cr
& \int {x{e^{1 - {x^2}}}} dx = \int {x{e^u}} \left( { - \frac{1}{{2x}}} \right)du \cr
& = \int {{e^u}} \left( { - \frac{1}{2}} \right)du \cr
& = - \frac{1}{2}\int {{e^u}du} \cr
& = - \frac{1}{2}{e^u} + C \cr
& {\text{Write in terms of }}x \cr
& = - \frac{1}{2}{e^{1 - {x^2}}} + C \cr} $$
