Answer
$2\pi^2$
Work Step by Step
Plug $p=\tan^{-1} x \implies dp=\dfrac{dv}{1+x^2}$
Re-arrange the given integral as follows:
$\int_{0}^{\infty} \dfrac{16 \tan^{-1} x dx}{1+x^2}=\int_{0}^{\pi/2} 16 p dp$ ....(1)
Then, we have
$\int_{0}^{\pi/2} 16 p dp=[8p^2]_{0}^{\pi/2}$
Thus, equation (1) becomes
$8(\dfrac{\pi}{2})^2=2\pi^2$
