How to prove $$\int_0^1 \left[ \frac{2}{\pi }\arctan \left(\frac 2 \pi \arctan \frac{1}{x} + \frac{1}{\pi }\ln \frac{1 + x}{1 - x}\right) - \frac{1}{2} \right]\frac{\mathrm{d}x} x = \frac{1}{2} \ln \left( \frac \pi {2\sqrt 2 } \right).$$
I have tried let $t=\frac1x$, but it seems no use! Could you help me to solve it?
One may show that this integral is equivalent to the integral in this problem as follows:
First, rewrite
$$\frac12 \log{\left ( \frac{1+x}{1-x} \right )} = \tanh^{-1}{x}$$
Then note that
$$\tan^{-1}{\left ( \frac1{x} \right )} = \frac{\pi}{2} - \tan^{-1}{x} $$
The integrand is then equal to
$$\left [\frac{2}{\pi} \tan^{-1}{\left (\frac{2}{\pi} \left (\frac{\pi}{2} - \tan^{-1}{x} + \tanh^{-1}{x} \right ) \right )}-\frac12 \right ] \frac1{x}$$
which is equal to
$$\left [\frac{2}{\pi} \tan^{-1}{\left (1+\frac{2}{\pi} \left ( \tanh^{-1}{x} - \tan^{-1}{x} \right ) \right )}-\frac12 \right ] \frac1{x}$$
Now, let
$$\tan^{-1}{(1+y)} = \frac{\pi}{4} + \tan^{-1}{w} $$
Then it is straight forward to show that
$$y = \frac{1+w}{1-w} - 1 = \frac{2 w}{1-w} \implies w=\frac{y}{y+2} $$
With $y=(2/\pi) (\tanh^{-1}{x} - \tan^{-1}{x} )$, we have
$$\frac{2}{\pi} \tan^{-1}{\left (1+\frac{2}{\pi} \left ( \tanh^{-1}{x} - \tan^{-1}{x} \right ) \right )}-\frac12 = \frac{2}{\pi} \tan^{-1}{\left (\frac{\tanh^{-1}{x} - \tan^{-1}{x} }{\tanh^{-1}{x} - \tan^{-1}{x} +\pi} \right )} $$
Thus, the integral in question is equal to
$$\frac{2}{\pi} \int_0^1 \frac{dx}{x} \tan^{-1}{\left (\frac{\tanh^{-1}{x} - \tan^{-1}{x} }{\tanh^{-1}{x} - \tan^{-1}{x} +\pi} \right )} $$
The evaluation of this integral is detailed in this arXiv paper.
