## Thomas' Calculus 13th Edition

$\ln 3$
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, $\ln |\dfrac{1-0}{1+0}|-\ln (\dfrac{1}{3}) =-\ln (\dfrac{1}{3}) =\ln 3$