Answer
The improper integral diverges.
Work Step by Step
We are given that $I=\int_{0}^{+\infty} \dfrac{2x}{x^2-1} \ dx$
Now, we have $\int_{0}^{+\infty} \dfrac{2x}{x^2-1} \ dx=\lim\limits_{a \to +\infty}\int_{0}^{a} \dfrac{2x}{x^2-1} \ dx$
or, $=\lim\limits_{a \to +\infty}[\ln |x^2-1|]_0^{a}$
or, $= \lim\limits_{a \to +\infty} [\ln |a^2-1|]-[\ln |(0)^2-1|]$
or, $= \lim\limits_{a \to +\infty} [\ln |a^2-1|]$
or, $=\ln |(\infty)^2-1|$
or, $=\infty$
Therefore, the given improper integral diverges.
