Answer
$$\ln \left(1+\frac{\pi}{2}\right)$$
Work Step by Step
\begin{align*}
\int_{0}^{\infty} \frac{d v}{\left(1+v^{2}\right)\left(1+\tan ^{-1} v\right)}&=\int_{0}^{\infty} \frac{\frac{1}{\left(1+v^{2}\right)}d v}{\left(1+\tan ^{-1} v\right)}\\
&=\lim _{b \rightarrow \infty}\left[\ln \left|1+\tan ^{-1} v\right|\right]_{0}^{b}\\
&=\lim _{b \rightarrow \infty}\left[\ln \left|1+\tan ^{-1} b\right|-\ln \left|1+\tan ^{-1} 0\right|\right]\\
&=\ln \left(1+\frac{\pi}{2}\right)-\ln (1+0)\\
&=\ln \left(1+\frac{\pi}{2}\right)
\end{align*}
