Answer
The improper integral diverges.
Work Step by Step
We are given that $I=\int_{-\infty}^{0} \dfrac{2x}{x^2-1} \ dx$
Now, we have $\int_{-\infty}^{0} \dfrac{2x}{x^2-1} \ dx=\lim\limits_{a \to -\infty}\int_{a}^{0} \dfrac{2x}{x^2-1} \ dx$
or, $=\lim\limits_{a \to -\infty}[\ln |x^2-1|]_a^{0}$
or, $= \lim\limits_{a \to -\infty} [\ln |0^2-1|]-[\ln |(a)^2-1|]$
or, $= -\lim\limits_{a \to -\infty} [\ln |a^2-1|]$
or, $=-\ln |(-\infty)^2-1|$
or, $=\infty$
Therefore, the given improper integral diverges.
