[passer-by]

Another version :

Define $ I= \int_{\frac{1}{a}}^a \frac{\arctan(x)}{x} \,\, dx $

Let $ u =\frac{1}{x}$ and substitute it in given integral.

Finally we obtain

$ I= \int_{\frac{1}{a}}^a \frac{\arctan(\frac{1}{u})}{u} \,\, du = \int_{\frac{1}{a}}^a \frac{\frac{\pi}{2}-\arctan(u)}{u} \,\, du = \pi \ln a - I $

Hence $ I= \frac{\pi \ln a}{2} $