Answer
$\ln 3$
Work Step by Step
Consider $f(t)= \int_2^{\infty} \dfrac{2 dt}{t^2-1}$
This can be re-written as:
$\int_2^{\infty} \dfrac{2 dt}{(t^2-1)}=\int_2^{\infty} \dfrac{2 dt}{(t-1)(t+1)}$
Now, $\int_2^{\infty} \dfrac{1}{(t-1)} dt-\int_2^{\infty} \dfrac{1}{(t+1)} dt=\lim\limits_{k \to \infty} [\ln |\dfrac{t-1}{t+1}|]_2^{k} $
and $ \lim\limits_{k\to \infty} [\ln |\dfrac{1-\dfrac{1}{k}}{1+\dfrac{1}{k}}|]-\ln (\dfrac{1}{3}) $
or, $\ln |\dfrac{1-0}{1+0}|-\ln (\dfrac{1}{3}) =-\ln (\dfrac{1}{3}) =\ln 3$
