Answer
$$\frac{\pi}{2} \ln \pi$$
Work Step by Step
$$
\begin{aligned}
\int_{0}^{\infty} \frac{\arctan \pi x-\arctan x}{x} d x &=\int_{0}^{\infty}\left[\frac{\arctan y x}{x}\right]_{y=1}^{y=\pi} d x \\
& =\int_{0}^{\infty} \int_{1}^{\pi} \frac{1}{1+y^{2} x^{2}} d y d x \\
&=\int_{1}^{\pi} \int_{0}^{\infty} \frac{1}{1+y^{2} x^{2}} d x d y \\
&=\int_{1}^{x} \lim _{t \rightarrow \infty}\left[\frac{\arctan y x}{y}\right]_{x=0}^{x-t} d y \\
&=\int_{1}^{\pi} \frac{\pi}{2 y} d y \\
&=\frac{\pi}{2}[\ln y]_{1}^{\pi} \\
&=\frac{\pi}{2} \ln \pi
\end{aligned}
$$
