Answer
$${\text{diverges}}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^{ - 1} {\frac{1}{{{x^{1/3}}}}} dx \cr
& {\text{Using the definition of improper integrals see page 1053}} \cr
& \underbrace {\int_{ - \infty }^b {f\left( x \right)dx = \mathop {\lim }\limits_{M \to - \infty } } \int_M^b {f\left( x \right)} dx}_ \Downarrow \cr
& \int_{ - \infty }^{ - 1} {\frac{1}{{{x^{1/3}}}}} dx = \mathop {\lim }\limits_{M \to - \infty } \int_M^{ - 1} {\frac{1}{{{x^{1/3}}}}} dx \cr
& = \mathop {\lim }\limits_{M \to - \infty } \int_M^{ - 1} {{x^{ - 1/3}}} dx \cr
& {\text{Integrating by the power rule}} \cr
& = \mathop {\lim }\limits_{M \to - \infty } \left[ {\frac{{{x^{2/3}}}}{{2/3}}} \right]_M^{ - 1} \cr
& = \frac{3}{2}\mathop {\lim }\limits_{M \to - \infty } \left[ {{x^{2/3}}} \right]_M^{ - 1} \cr
& = \frac{3}{2}\mathop {\lim }\limits_{M \to - \infty } \left[ {{{\left( { - 1} \right)}^{2/3}} - {M^{2/3}}} \right] \cr
& = \frac{3}{2}\mathop {\lim }\limits_{M \to - \infty } \left[ {1 - {M^{2/3}}} \right] \cr
& {\text{Evaluate the limit when }}M \to - \infty \cr
& = \frac{3}{2}\left[ {1 - {{\left( { - \infty } \right)}^{2/3}}} \right] \cr
& = \frac{3}{2}\left[ {1 - \infty } \right] \cr
& = - \infty \cr
& {\text{diverges}} \cr} $$
