Answer
$\ln 3$
Work Step by Step
Since, we have $ \int_2^{\infty} \dfrac{2 dt}{t^2-1}=\int_2^{\infty} \dfrac{2 dt}{(t-1)(t+1)}$
or, $=\int_2^{\infty} \dfrac{1}{(t-1)}-\int_2^{\infty} \dfrac{1t}{(t+1)} dt$
or, $=\lim\limits_{a \to \infty} [\ln |\dfrac{t-1}{t+1}|]_2^{\infty} $
and $ \lim\limits_{a \to \infty} [\ln |\dfrac{t-1}{t+1}|]_2^{\infty}=\lim\limits_{a \to \infty} [\ln |\dfrac{1-1/t}{1+1/t}|]_2^{\infty}-\ln (1/3) $
or, $=\ln 3$
