How to prove ∫10tan−1[tanh−1x−tan−1xπ+tanh−1x−tan−1x]dxx=π8lnπ28?\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}?

How can one prove that
$$\int_0^1 \tan^{-1}\left[\frac{\tanh^{-1}x-\tan^{-1}x}{\pi+\tanh^{-1}x-\tan^{-1}x}\right]\frac{dx}{x}=\frac{\pi}{8}\ln\frac{\pi^2}{8}?$$

Answer

Attribution
Source : Link , Question Author : larry , Answer Author : Community