Answer
$\displaystyle \frac{\ln 5}{4}$
(converges)
Work Step by Step
$I=\displaystyle \int_{2}^{\infty}\frac{dv}{v^{2}+2v-3}=\lim_{t\rightarrow\infty}\int_{2}^{t}\frac{dv}{(v+3)(v-1)}=$
$\displaystyle \frac{1}{(v+3)(v-1)}=\frac{A}{v+3}+\frac{B}{v-1}$
$\displaystyle \frac{1}{(v+3)(v-1)}=\frac{A(v-1)+B(v+3)}{v+3}$
$\displaystyle \frac{1}{(v+3)(v-1)}=\frac{(A+B)v+(-A+3B)}{v+3}$
$\Rightarrow\left\{\begin{array}{l}
A+B=0\\
-A+3B=1
\end{array}\right.$
Adding, $4B=1\Rightarrow B=1/4,\quad A=-1/4$
$\displaystyle \frac{1}{(v+3)(v-1)}=\frac{-1/4}{v+3}+\frac{1/4}{v-1}$
$I=\displaystyle \lim_{t\rightarrow\infty}\left[-\frac{1}{4}\ln|v+3|+\frac{1}{4}\ln|v-1|\right]_{2}^{t}$
$=\displaystyle \frac{1}{4}\lim_{t\rightarrow\infty}\left[\ln\frac{v-1}{v+3}\right]_{2}^{t}$
$=\displaystyle \frac{1}{4}\lim_{t\rightarrow\infty}\left(\ln\frac{t-1}{t+3}-\ln\frac{1}{5}\right)$
$=\displaystyle \frac{1}{4}\lim_{t\rightarrow\infty}\left(\ln\frac{\frac{t-1}{t+3}}{1/5}\right)$
$=\displaystyle \frac{1}{4}\ln[\lim_{t\rightarrow\infty}\frac{5t-5}{t+3}]$
$=\displaystyle \frac{1}{4}(\ln 5)$
$=\displaystyle \frac{\ln 5}{4}$
(converges)
