Answer
$${\text{diverges}}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^\infty {\frac{x}{{\sqrt {{x^2} + 2} }}} dx \cr
& {\text{Use the definition }}\int_{ - \infty }^\infty {f\left( x \right)} dx = \int_{ - \infty }^c {f\left( x \right)} dx + \int_c^{ + \infty } {f\left( x \right)} dx \cr
& {\text{Let }}c = 0. \cr
& \int_{ - \infty }^\infty {\frac{x}{{\sqrt {{x^2} + 2} }}} dx = \int_{ - \infty }^0 {\frac{x}{{\sqrt {{x^2} + 2} }}} dx + \int_0^{ + \infty } {\frac{x}{{\sqrt {{x^2} + 2} }}} dx \cr
& {\text{Evaluating the integrals on the right side separately}} \cr
& \int_{ - \infty }^0 {\frac{x}{{\sqrt {{x^2} + 2} }}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{x}{{\sqrt {{x^2} + 2} }}} dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{a \to - \infty } \left[ {\sqrt {{x^2} + 2} } \right]_a^0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{a \to - \infty } \left[ {\sqrt {{0^2} + 2} - \sqrt {{a^2} + 2} } \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop {\lim }\limits_{a \to - \infty } \left[ {\sqrt 2 - \sqrt {{a^2} + 2} } \right] \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Find the limit when }}a \to - \infty \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 2 - \sqrt {{{\left( \infty \right)}^2} + 2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \sqrt 2 - \infty \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \infty \cr
& {\text{The integral }}\int_{ - \infty }^0 {\frac{x}{{\sqrt {{x^2} + 2} }}} dx{\text{ diverges, so}}{\text{ }} \cr
& {\text{the given integral }}\int_{ - \infty }^\infty {\frac{x}{{\sqrt {{x^2} + 2} }}} dx{\text{ diverges}} \cr} $$
