Answer
$\ln 3$
Work Step by Step
Let $f(x)= \int_{-\infty}^{-2} \dfrac{2}{x^2-1} dx$
Now, $\lim\limits_{k \to -\infty} f(x)= \lim\limits_{k \to -\infty}\int_{k}^{-2} [2(\dfrac{1}{x^2-1} dx)]=\lim\limits_{k \to -\infty}\int_{k}^{-2} \dfrac{-1}{x+1}+\dfrac{1}{x-1} $
Thus, $\lim\limits_{k \to -\infty} [\ln |x-1|-\ln |x+1|]_{(k)}^{-2}=\lim\limits_{k \to -\infty} \ln 3 - \lim\limits_{k \to -\infty} \ln |\dfrac{k-1}{k+1}|=\ln 3-0=\ln 3$
