Show that \(artanh (x) = \frac{1}{2}\ln \left( \frac{1+x}{1-x} \right)\,, \quad |x| \lt 1 \)

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\begin{align} Let\ y & = artanh(x) \\ x & = tanh(y)\\ \\ x & = \frac{e^y -e^{-y}}{e^y+e^{-y}} \\ \\ x & = \frac{e^{2y} -e^{0}}{e^{2y}+e^{0}}\\ \\ e^{2y}-1 & = xe^{2y}+x \\ 0 & = e^{2y}-xe^{2y} -x -1 \\ 0 & = e^{2y}(1-x) - (1+x) \\ e^{2y} & = \frac{1+x}{1-x} \\ 2y & = \ln\left(\frac{1+x}{1-x} \right) \\ y & = \frac{1}{2} \ln\left(\frac{1+x}{1-x} \right) \\ \implies artanh (x) & = \frac{1}{2} \ln\left(\frac{1+x}{1-x} \right)\ , \quad |x| \lt 1 \end{align}