Problem with Integral attempt

$$\begin{eqnarray*}\int_{0}^{+\infty}\frac{\arctan(\pi x)-\arctan x}{x}\,dx &=& \int_{0}^{+\infty}\frac{1}{x}\int_{1}^{\pi}\frac{x}{1+a^2 x^2}\,da\,dx\\&=&\int_{1}^{\pi}\frac{\pi}{2a}\,da\\&=&\color{red}{\frac{\pi}{2}\log\pi}.\end{eqnarray*}$$