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   <subfield code="a">As an application of Cauchy's Theorem we prove that $$\int_{0}^{1}{\rm arctan}\left(\frac{{\rm arctanh} x-{\rm arctan} x}{\pi+{\rm arctanh} x-{\rm arctan} x}\right) \frac{dx}{x}= \frac{\pi}{8}{\rm log}\frac{\pi^{2}}{8}$$ ∫ 0 1 arctan arctanh x - arctan x π + arctanh x - arctan x d x x = π 8 log π 2 8 answering by this means a question posted in 1984 by J. A. Morrison in the Problem Section of the journal SIAM Review.</subfield>
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