`tanh^-1 x = 1/2 ln((1+x)/(1-x)) , -1

Given to prove

$\displaystyle{{\tanh}}^{{-{1}}}{x}=\frac{{1}}{{2}}{\ln{{\left(\frac{{{1}+{x}}}{{{1}-{x}}}\right)}}}$

so let

$\displaystyle{{\tanh}}^{{-{1}}}{x}={y}$

=> $\displaystyle{x}={\tanh{{\left({y}\right)}}}$

$\displaystyle{x}=\frac{{{{e}}^{{y}}-{{e}}^{{-{{y}}}}}}{{{{e}}^{{y}}+{{e}}^{{-{{y}}}}}}$

=> $\displaystyle{\left({{e}}^{{y}}+{{e}}^{{-{{y}}}}\right)}\cdot{x}={\left({{e}}^{{y}}-{{e}}^{{-{{y}}}}\right)}$

=> $\displaystyle{x}{{e}}^{{y}}+{x}{{e}}^{{-{{y}}}}={{e}}^{{y}}-{{e}}^{{-{{y}}}}$

=> $\displaystyle\frac{{{x}{{e}}^{{{2}{y}}}+{x}}}{{{e}}^{{y}}}=\frac{{{{e}}^{{2}}{y}-{1}}}{{{e}}^{{y}}}$

=> $\displaystyle{\left({x}{{e}}^{{{2}{y}}}+{x}\right)}={\left({{e}}^{{{2}{y}}}-{1}\right)}$

=> $\displaystyle{\left({x}{{e}}^{{{2}{y}}}+{x}\right)}-{{e}}^{{{2}{y}}}+{1}={0}$

=> $\displaystyle{{e}}^{{{2}{y}}}{\left({x}-{1}\right)}+{x}+{1}={0}$

=> $\displaystyle{\left({x}-{1}\right)}{\left({{e}}^{{{2}{y}}}\right)}=-{\left({x}+{1}\right)}$

=> $\displaystyle{{e}}^{{{2}{y}}}=-\frac{{{x}+{1}}}{{{x}-{1}}}$

=> $\displaystyle{{e}}^{{{2}{y}}}=\frac{{{1}+{x}}}{{{1}-{x}}}$

=> $\displaystyle{{e}}^{{{2}{y}}}={\left(\frac{{{1}+{x}}}{{{1}-{x}}}\right)}$

=> $\displaystyle{2}{y}={\ln{{\left({\left(\frac{{{1}+{x}}}{{{1}-{x}}}\right)}\right)}}}$

=> $\displaystyle{y}=\frac{{1}}{{2}}{\ln{{\left({\left(\frac{{{1}+{x}}}{{{1}-{x}}}\right)}\right)}}}$

so,

$\displaystyle{{\tanh}}^{{-{1}}}{x}=\frac{{1}}{{2}}{\ln{{\left(\frac{{{1}+{x}}}{{{1}-{x}}}\right)}}}$

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