Problem
Prove formula $\operatorname{arctanh} x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)$
Attempt to solve
To start off with definition of functions $\sinh(x)$ and $\cosh(x)$
$$ \cosh(x)=\frac{e^x+e^{-x}}{2} $$
$$ \sinh(x) = \frac{e^x-e^{-x}}{2} $$
Hyperbolic tangent is defined as:
$$ \tanh(x)=\frac{\sinh(x)}{\cosh(x)}=\frac{e^x-e^{-x}}{e^{x}+e^{-x}} $$
Notation $\text{arctanh}(x)$ means area tangent which is inverse of hyperbolic tangent.
$$ \operatorname{arctanh}(x)=\tanh^{-1}(x) $$
Trying to invert the $\tanh(x)$ we get:
$$\begin{align}
\frac{e^x-e^{-x}}{e^x+e^{-x}} &= y &\implies\\
e^x-e^{-x}&=y(e^x+e^{-x}) &\implies\\
e^x-e^{-x}&=ye^{x}+ye^{-x} &\implies\\
e^x(1-y)&=e^{-x}(1+y) &\implies\\
\ln(e^x(1-y)) &= \ln(e^{-x}(1+y)) &\implies\\
\ln(e^x)+\ln(1-y) &= \ln(e^{-x})+\ln(1+y) &\implies\\
x + \ln(1-y) &= -x + \ln(1+y) &\implies\\
2x &= \ln(1+y)-\ln(1-y)&\implies\\
x &= \frac{1}{2} \ln \frac{1+y}{1-y}
\end{align}
$$
By switching variables we get:
$$ \implies \operatorname{arctanh}(y) = \frac{1}{2} \ln \left(\frac{1+y}{1-y}\right) $$
Best Answer
Looks fine to me. Well done!
Just a small thing, a matter of taste: I'd prefer to do this:
\begin{align*} e^x(1-y) &= e^{-x}(1+y)\\ e^{2x} &=\dfrac{1+y}{1-y}\\ x &= \dfrac{1}{2}\ln\left(\dfrac{1+y}{1-y}\right). \end{align*}