Computation of a Definite Integral

Abstract

As an application of Cauchy's Theorem we prove that $\int_0^1\arctan\left({\arctanh x-\arctan x\over \pi+\arctanh x-\arctan x}\right) {dx\over x}= {\pi\over 8}\log{\pi^2\over 8}$ answering a question first posted in Mathematics Stack Exchange and then in MathOverflow.