show that tanh^-1x=ln((1+x)/(1-x))/2

Answer 1

y = tanh^-1(x)
x = tanh y = (e^y - e^-1)/(e^y + e^-y) = (e^(2y)-1)/(e^(2y)+1)
xe^(2y) + x = e^(2y) - 1
e^(2y) = (1+x)/(1-x)
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