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 P: 134 then we have from trigonometry that: $$Artanh(ix)=iartan(x)$$ On the other hand... $$2artanh(x)=log(1-x)-log(1+x)$$ $$2artanh(ix)=log(1-ix)-log(1+ix)$$ $$2artanh(i/x)=log(1-i/x)-log(1+i/x)$$ then $$artanh(ix)+artanh(i/x)=i(artan(x)+artan(1/x))$$ taking the sum of all the logs you have.. $$-log(1+ix)-log(1+i/x)+log(1-ix)+log(1-i/x)$$ $$-log(2i/x)+log(-2i/x)=log(-1)=i\pi$$ - i have taken the first "branch" of log ..(the angle goes from 0 to 360 degrees), the factor "2" comes from the definition of artanh(x) in the form of log (log in basis e)..