Answer
$${\text{The integral is divergent}}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^0 {\frac{x}{{{x^2} + 3}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{,}} \cr
& \int_{ - \infty }^0 {\frac{x}{{{x^2} + 3}}} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{x}{{{x^2} + 3}}} dx \cr
& \cr
& {\text{integrating }}\int {\frac{x}{{{x^2} + 3}}} dx \cr
& {\text{set }}u = {x^2} + 3{\text{ then }}\frac{{du}}{{dx}} = 2x,\,\,\,\,\frac{{du}}{{2x}} = dx \cr
& \int {\frac{x}{{{x^2} + 3}}} dx = \int {\frac{x}{u}} \left( {\frac{{du}}{{2x}}} \right) = \frac{1}{2}\int {\frac{{du}}{u}} \cr
& = \frac{1}{2}\ln \left| u \right| + C \cr
& {\text{replace }}u = {x^2} + 3 \cr
& = \frac{1}{2}\ln \left( {{x^2} + 3} \right) + C \cr
& {\text{then}} \cr
& \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{x}{{{x^2} + 3}}} dx = \mathop {\lim }\limits_{a \to - \infty } \left( {\frac{1}{2}\ln \left( {{x^2} + 3} \right)} \right)_a^0 \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left( {\ln \left( {{x^2} + 3} \right)} \right)_a^0 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\left( {{\text{see page 388}}} \right) \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left( {\ln \left( {{0^2} + 3} \right) - \ln \left( {{a^2} + 3} \right)} \right) \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left( {\ln \left( 3 \right) - \ln \left( {{a^2} + 3} \right)} \right) \cr
& {\text{evaluating the limit when }}a \to - \infty \cr
& = \frac{1}{2}\left( {\ln \left( 3 \right) - \ln \left( \infty \right)} \right) \cr
& = \infty \cr
& then \cr
& {\text{The integral is divergent}} \cr} $$
