Answer
$$2 \pi^{2} $$
Work Step by Step
\begin{align*}
\int_{0}^{\infty} \frac{16 \tan ^{-1} x}{1+x^{2}} d x&= 16\int_{0}^{\infty} \frac{1}{1+x^{2}} \tan ^{-1} x dx\\
&=\lim _{b \rightarrow \infty}\left[8\left(\tan ^{-1} x\right)^{2}\right]_{0}^{b}\\
&=\lim _{b \rightarrow \infty}\left[8\left(\tan ^{-1} b\right)^{2}-8\left(\tan ^{-1} 0\right)^{2}\right]\\&=8\left(\frac{\pi}{2}\right)^{2}-8(0)\\
&=2 \pi^{2}
\end{align*}