## Calculus: Early Transcendentals 8th Edition

$\displaystyle \frac{\ln 5}{4}$ (converges)
$I=\displaystyle \int_{2}^{\infty}\frac{dv}{v^{2}+2v-3}=\lim_{t\rightarrow\infty}\int_{2}^{t}\frac{dv}{(v+3)(v-1)}=$ $\displaystyle \frac{1}{(v+3)(v-1)}=\frac{A}{v+3}+\frac{B}{v-1}$ $\displaystyle \frac{1}{(v+3)(v-1)}=\frac{A(v-1)+B(v+3)}{v+3}$ $\displaystyle \frac{1}{(v+3)(v-1)}=\frac{(A+B)v+(-A+3B)}{v+3}$ $\Rightarrow\left\{\begin{array}{l} A+B=0\\ -A+3B=1 \end{array}\right.$ 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)