Converting $\tanh^{-1}{x}$ to an expression involving the natural logarithm I know how to convert $\sinh^{-1}{x}$ and $\cosh^{-1}{x}$ to $\ln{|x+\sqrt{x^2 \pm 1}|}$, but for some reason I am struggling to do the same for the following statement: 
$$\tanh^{-1}{\frac{x}{2}}$$
Can someone please show me how to convert it to a $\ln$ form? thanks!

$$\tanh x=\frac{e^{2x}-1}{e^{2x}+1} \Rightarrow \tanh \frac{x}{2}=\frac{e^{x}-1}{e^{x}+1} $$  In order to find $artanh\frac{x}{2}$ we have to solve following equation :
$$\frac{x}{2}= \frac {e^{2artanh \frac{x}{2}}-1}{e^{2artanh \frac{x}{2}}+1} \Rightarrow x \cdot e^{2artanh \frac{x}{2}}+x=2e^{2artanh \frac{x}{2}}-2 \Rightarrow$$
$$x+2=(2-x) \cdot e^{2artanh \frac{x}{2}} \Rightarrow e^{2artanh \frac{x}{2}}=\frac{2+x}{2-x} \Rightarrow artanh \frac{x}{2} =\frac{1}{2} \ln \left(\frac{2+x}{2-x}\right) ; |x| < 2$$
