The following identity on MATH.SE

$$\int_0^{1}\arctan\left(\frac{\mathrm{arctanh}\ x-\arctan{x}}{\pi+\mathrm{arctanh}\ x-\arctan{x}}\right)\frac{dx}{x}=\frac{\pi}{8}\log\frac{\pi^2}{8}$$

seems to be very difficult to prove.

Question: I worked on this identity for several days without any success. Is there any clue how to prove this integral identity?

The following identity on MathSE

$$\int_0^{1}\arctan\left(\frac{\mathrm{arctanh}\ x-\arctan{x}}{\pi+\mathrm{arctanh}\ x-\arctan{x}}\right)\frac{dx}{x}=\frac{\pi}{8}\log\frac{\pi^2}{8}$$

seems to be very difficult to prove.

Question: I worked on this identity for several days without any success. Is there any clue how to prove this integral identity?

The following identity on MATH.SE

$$\int_0^{1}\arctan\left(\frac{\mathrm{arctanh}\ x-\arctan{x}}{\pi+\mathrm{arctanh}\ x-\arctan{x}}\right)\frac{dx}{x}=\frac{\pi}{8}\log\frac{\pi^2}{8}$$

seems to be very difficult to prove.

Question: I worked on this identity for several days without any success. Is there any clue how to prove this integral identity?

The following identity on MATH.SE

$$\int_0^{1}\arctan\left(\frac{\mathrm{arctanh}\ x-\arctan{x}}{\pi+\mathrm{arctanh}\ x-\arctan{x}}\right)\frac{dx}{x}=\frac{\pi}{8}\log\frac{\pi^2}{8}$$

seems to be very difficult to prove.

Question: I worked on this identity for several days without any success. Is there any clue how to prove this integral identity?