Answer
$${\text{Diverges}}$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {\frac{x}{{{{\left( {{x^2} - 1} \right)}^{5/3}}}}} dx \cr
& {\text{The integrand }}\frac{x}{{{{\left( {{x^2} - 1} \right)}^{5/3}}}}{\text{ is not continuous to }}x = \pm 1,{\text{ then}} \cr
& {\text{Improper Integral with an Infinite Limit of Integration}} \cr
& \int_{ - 1}^1 {\frac{x}{{{{\left( {{x^2} - 1} \right)}^{5/3}}}}} dx = \mathop {\lim }\limits_{a \to - {1^ + }} \int_{ - 1}^0 {\frac{x}{{{{\left( {{x^2} - 1} \right)}^{5/3}}}}} dx + \mathop {\lim }\limits_{b \to {1^ - }} \int_0^b {\frac{x}{{{{\left( {{x^2} - 1} \right)}^{5/3}}}}} dx \cr
& {\text{*Computing }}\mathop {\lim }\limits_{a \to - {1^ + }} \int_{ - 1}^0 {\frac{x}{{{{\left( {{x^2} - 1} \right)}^{5/3}}}}} dx \cr
& \mathop {\lim }\limits_{a \to - {1^ + }} \int_{ - 1}^0 {\frac{x}{{{{\left( {{x^2} - 1} \right)}^{5/3}}}}} dx = \frac{1}{2}\left( { - \frac{3}{2}} \right)\mathop {\lim }\limits_{a \to - {1^ + }} \left[ {\frac{1}{{{{\left( {{x^2} - 1} \right)}^{2/3}}}}} \right]_a^0 \cr
& = - \frac{3}{4}\mathop {\lim }\limits_{a \to - {1^ + }} \left[ {\frac{1}{{{{\left( {{x^2} - 1} \right)}^{2/3}}}}} \right]_a^0 \cr
& = - \frac{3}{4}\mathop {\lim }\limits_{a \to - {1^ + }} \left[ {\frac{1}{{{{\left( {{0^2} - 1} \right)}^{2/3}}}} - \frac{1}{{{{\left( {{a^2} - 1} \right)}^{2/3}}}}} \right] \cr
& = - \frac{3}{4}\mathop {\lim }\limits_{a \to - {1^ + }} \left[ {1 - \frac{1}{{{{\left( {{a^2} - 1} \right)}^{2/3}}}}} \right] \cr
& = - \frac{3}{4}\left[ {1 - \frac{1}{{{{\left( {{{\left( { - 1} \right)}^2} - 1} \right)}^{2/3}}}}} \right] \cr
& = - \infty \cr
& {\text{Diverges to }}- \infty {\text{, There is no need now to check }} \cr
& \mathop {\lim }\limits_{b \to {1^ - }} \int_0^b {\frac{x}{{{{\left( {{x^2} - 1} \right)}^{5/3}}}}} dx{\text{ because one of two pieces of the}} \cr
& {\text{ integral diverges, then}} \cr
& \int_{ - 1}^1 {\frac{x}{{{{\left( {{x^2} - 1} \right)}^{5/3}}}}} dx{\text{ diverges}} \cr} $$
