Answer
$\ln 3$
Work Step by Step
Consider $f(x)= \int_{-\infty}^{-2} \dfrac{2}{x^2-1} dx$
Since, we have $\lim\limits_{a \to -\infty} f(x)= \lim\limits_{a \to -\infty}\int_{a}^{-2} [2(\dfrac{1}{x^2-1} dx)]=\lim\limits_{a \to -\infty}\int_{a}^{-2} \dfrac{-1}{x+1}+\dfrac{1}{x-1} $
or, $=\lim\limits_{a \to -\infty} [\ln |x-1|-\ln |x+1|]_{a}^{-2}$
or, $=\lim\limits_{a \to -\infty} \ln 3 -\ln |\dfrac{a-1}{a+1}|$
or, $=\ln 3$
