Answer
Diverges
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^{ - 3} {\frac{x}{{4 - {x^2}}}} dx \cr
& {\text{Using the definition of improper integrals }} \cr
& \underbrace {\int_{ - \infty }^b {f\left( x \right)dx = \mathop {\lim }\limits_{t \to - \infty } } \int_t^b {f\left( x \right)} dx}_ \Downarrow \cr
& \int_{ - \infty }^{ - 3} {\frac{x}{{4 - {x^2}}}} dx = \mathop {\lim }\limits_{t \to - \infty } \int_t^{ - 3} {\frac{x}{{4 - {x^2}}}} dx \cr
& {\text{Rewrite the integrand}} \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{t \to - \infty } \int_t^{ - 3} {\frac{{ - 2x}}{{4 - {x^2}}}} dx \cr
& {\text{Integrate}} \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{t \to - \infty } \left[ {\ln \left| {4 - {x^2}} \right|} \right]_t^{ - 3} \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{t \to - \infty } \left[ {\ln \left| {4 - {{\left( { - 3} \right)}^2}} \right| - \ln \left| {4 - {t^2}} \right|} \right] \cr
& = - \frac{1}{2}\mathop {\lim }\limits_{t \to - \infty } \left[ {\ln \left( 5 \right) - \ln \left| {4 - {t^2}} \right|} \right] \cr
& {\text{Evaluate the limit when }}t \to - \infty \cr
& = - \frac{1}{2}\left[ {\ln \left( 5 \right) - \infty } \right] \cr
& = \infty \cr
& {\text{Therefore}}{\text{, the integral diverges.}} \cr} $$
