Answer
$\ln |1+\dfrac{\pi}{2}|$
Work Step by Step
Plug $p=1+\tan^{-1} v \implies dp=\dfrac{dv}{1+v^2}$
Re-arrange the given integral as follows:
$\int_{0}^{\infty} \dfrac{dv}{(1+v^2)(1+\tan^{-1} v)}=\int_{1}^{1+\pi/2} \dfrac{dp}{(p)}$ ....(1)
Then, we have
$\int_{1}^{1+\pi/2} \dfrac{dp}{(p)}=[\ln p]_{1}^{1+\pi/2}$
Thus, equation (1) becomes
$\ln |1+\dfrac{\pi}{2}|-\ln |1|=\ln |1+\dfrac{\pi}{2}|$
