Answer
$\ln 4$
Work Step by Step
Consider $f(v)=\int_2^{\infty} \dfrac{2}{v^2-v}$
This can be written as:
$ \int_2^{\infty} \dfrac{2}{v^2-v}=\lim\limits_{k \to \infty} \int_2^{k}\dfrac{2}{v-1}-\dfrac{2}{v} dv$
Now $2[ \lim\limits_{k \to \infty} \ln |k-1| -\lim\limits_{k \to \infty} \ln (k)]- 2[ \ln (1)-\ln 2]=2 \ln 2$
Thus, $ 2\ln 2=\ln (2^2) =\ln 4$
