Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^{ + \infty } {x{e^{1 - {x^2}}}} dx \cr
& {\text{Improper Integral with an Infinite Limit of Integration}} \cr
& \int_{ - \infty }^{ + \infty } {x{e^{1 - {x^2}}}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_b^0 {x{e^{1 - {x^2}}}} dx + \mathop {\lim }\limits_{b \to + \infty } \int_0^b {x{e^{1 - {x^2}}}} dx \cr
& {\text{Integrating}} \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {{e^{1 - {x^2}}}} \right]_a^0 - \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {{e^{1 - {x^2}}}} \right]_0^b \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {{e^{1 - {{\left( 0 \right)}^2}}} - {e^{1 - {{\left( a \right)}^2}}}} \right] - \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {{e^{1 - {{\left( b \right)}^2}}} - {e^{1 - {{\left( 0 \right)}^2}}}} \right] \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {e - {e^{1 - {{\left( a \right)}^2}}}} \right] - \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {{e^{1 - {{\left( b \right)}^2}}} - e} \right] \cr
& {\text{Evaluate the limits}} \cr
& = - \frac{1}{2}\left[ {e - {e^{1 - {{\left( { - \infty } \right)}^2}}}} \right] - \frac{1}{2}\left[ {{e^{1 - {{\left( { + \infty } \right)}^2}}} - e} \right] \cr
& = - \frac{1}{2}\left[ {e - 0} \right] - \frac{1}{2}\left[ {0 - e} \right] \cr
& = - \frac{1}{2}e + \frac{1}{2}e \cr
& = 0 \cr} $$
