How do you show that #tanh^-1(x) = (1/2) ln ((1+x)/(1-x))#?

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How do you show that #tanh^-1(x) = (1/2) ln ((1+x)/(1-x))#?

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Answer 1 · 2016-05-07T04:52:21

Let #y=tanh^(-1)x#. Then # x = tanh y = (e^y-e^(-y))/(e^y+e^(-y))=(e^(2y)-1)/(e^(2y)+1)#.. Solving for #e^(2y)# and then for y, by equating logarithms, we get the logarithmic form.

Explanation:

Note that I had corrected the question by changing 'x-1' to '1-x'.

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How do you show that #tanh^-1(x) = (1/2) ln ((1+x)/(1-x))#?

1 Answer

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