Inspired by Rishabh Cool

Let us define $y=g(x)=\tanh(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$ The inverse function is $g^{-1}(x)=arctanh(x)$ Another way to express $arctanh(x)$ is:

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

This means that all three of the functions are equations for $arctanh(x)$ .

The maclaurin series for $\ln(1-x)$ is

$-\sum_{n=1}^{\infty}\frac{x^{n}}{n}$ .

The equations above for $arctanh(x)$ are the same as $\frac{1}{2}(\ln(1+x)-\ln(1-x))$ using the laws of logarithms that $\ln(\frac{a}{b})=\ln(a)-\ln(b)$ Replacing the logarithms with the maclaurin series for $-\ln(1-x)$ , we get

$\begin{aligned} arctanh(x)&=\frac{1}{2}(\sum_{n=1}^{\infty}\frac{x^{n}}{n}-\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{n}}{n})\\ &=\frac{1}{2}((\frac{x^{1}}{1}+\frac{x^{2}}{2}+\frac{x^{3}}{3}...)-(-\frac{x^{1}}{1}+\frac{x^{2}}{2}-\frac{x^{3}}{3}...))\\ &=\frac{1}{2}(2(\frac{x^{1}}{1}+\frac{x^{3}}{3}+\frac{x^{5}}{5}...))\\ &=\sum_{n=1}^{\infty}\frac{x^{2n-1}}{2n-1}\\ \end{aligned}$