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