Prove that $\displaystyle\tanh^{-1}(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ for $-1Prove that $\displaystyle\tanh^{-1}(x)=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ for $-1<x<1$.
So far I have got $y=\tanh^{-1}(x)\Longleftrightarrow\tanh(y)=x$ . 
Differentiating, $\displaystyle\tanh^{-1}(y)=x \Rightarrow \text{sech}^2(y)\cdot y'=1\Rightarrow y'=\frac{1}{\text{sech}^2(y)}=\frac{1}{\tanh^2(y)-1}=\frac{1}{x^2-1}$.
Stucked here.

The hyperbolic arctangent of $x$, $y=\operatorname{arctanh} x$, is the solution of $\tanh y=x$, i.e.
$$ \tanh y=\frac{e^y-e^{-y}}{e^y+e^{-y}}=\frac{e^{2y}-1}{e^{2y}+1}=x,\tag{1}$$
leading to:
$$ e^{2y} = \frac{x+1}{x-1}\tag{2}$$
so that:
$$\operatorname{arctanh} x = y = \frac{1}{2}\log e^{2y} = \frac{1}{2}\,\log\left(\frac{1+x}{1-x}\right).\tag{3}$$
As an alternative, since the hyperbolic arctangent is the inverse function of the hyperbolic tangent,
$$\frac{d}{dz}\operatorname{arctanh} z = \frac{1}{1-z^2} = \frac{1}{2}\left(\frac{1}{1+z}+\frac{1}{1-z}\right)=\frac{1}{2}\frac{d}{dz}\log\left(\frac{1+z}{1-z}\right) \tag{4}$$
and we just have to check that the RHS and LHS of $(3)$ match in $x=0$.
