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