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Question

# Prove that $\tanh ^{-1} x=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right), \quad-1.

Solution

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First let's substitute as follows,

$\begin{array}{lcl} u&=&\tanh^{-1}x\\ \tanh u&=& x \end{array}$

Now rewriting the first part,

\begin{aligned} \frac{\sinh u}{\cosh u}&=x\\ \frac{e^u-e^{-u}}{e^u+e^{-u}}&=x\\ \end{aligned}

Simplifying,

\begin{aligned} e^u-e^{-u}&=x(e^u+e^{-u})\\ e^{2u}-1&=x(e^{2u}+1)\\ e^{2u}-1&=xe^{2u}+x\\ e^{2u}-xe^{2u}&=1+x\\ e^{2u}(1-x)&=1+x\\ e^{2u}&=\frac{1+x}{1-x} \end{aligned}

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