Answer
$ - \frac{1}{4}$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^0 {\frac{x}{{{{\left( {{x^2} + 1} \right)}^3}}}} dx \cr
& {\text{Using the definition of improper integrals }} \cr
& \underbrace {\int_{ - \infty }^b {f\left( x \right)dx = \mathop {\lim }\limits_{t \to - \infty } } \int_t^b {f\left( x \right)} dx}_ \Downarrow \cr
& \int_{ - \infty }^0 {\frac{x}{{{{\left( {{x^2} + 1} \right)}^3}}}} dx = \mathop {\lim }\limits_{t \to - \infty } \int_t^0 {\frac{x}{{{{\left( {{x^2} + 1} \right)}^3}}}} dx \cr
& {\text{Rewrite the integrand}} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{t \to - \infty } \int_t^0 {{{\left( {{x^2} + 1} \right)}^{ - 3}}\left( {2x} \right)} dx \cr
& {\text{Integrate by using the power rule}} \cr
& = \frac{1}{2}\mathop {\lim }\limits_{t \to - \infty } \left[ {\frac{{{{\left( {{x^2} + 1} \right)}^{ - 2}}}}{{ - 2}}} \right]_t^0 \cr
& = - \frac{1}{4}\mathop {\lim }\limits_{t \to - \infty } \left[ {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}} \right]_t^0 \cr
& = - \frac{1}{4}\mathop {\lim }\limits_{t \to - \infty } \left[ {\frac{1}{{{{\left( {{0^2} + 1} \right)}^2}}} - \frac{1}{{{{\left( {{t^2} + 1} \right)}^2}}}} \right] \cr
& {\text{Evaluate the limit when }}t \to - \infty \cr
& = - \frac{1}{4}\mathop {\lim }\limits_{t \to - \infty } \left[ {\frac{1}{1} - 0} \right] \cr
& = - \frac{1}{4} \cr
& {\text{Therefore}}{\text{, the integral converges.}} \cr} $$
