Showing $\int_0^1\tan^{-1}\frac{f(x)}{1+f(x)}\frac{dx}x=\frac\pi8\ln\frac{\pi^2}8$, where $f(x)=\frac1\pi(\tanh^{-1}x-\tan^{-1}x)$, with real analysis

Show that: $I=\int_{0}^{1}\tan^{-1}\left(\frac{\tanh^{-1}x-\tan^{-1}x}{\pi+\tanh^{-1}x-\tan^{-1}x}\right)\frac{dx}{x}=\frac{\pi}{8}\ln\!\frac{\pi^{2}}{8}$

I found this question Here on MSE. Here is the solution obtained by means of complex analysis.

I found this question on MathOverflow Here, but no answers using purely real analysis were provided.