Answer
$\ln 4$
Work Step by Step
We rewrite the integral as:
$2 \int_2^{\infty} \dfrac{1}{v^2-v}=\lim\limits_{a \to \infty} \int_2^{a}\dfrac{1}{v-1}-\dfrac{1}{v} dv$
or, $=\lim\limits_{a \to \infty} 2( \ln |a-1| -\ln a)- 2( \ln 1-\ln 2)$
or, $=2 ln 2$
and $ 2\ln 2=\ln 2^2 =\ln 4$
