Htykuut

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!

Set $ y = tanh^{-1} t $ and take $tanh$ to take both sides so we have $$ tanh y = t .$$

Now convert the $tanh$ term into it's definition in terms of exponentials: $$ t = frac{ e^{y} - e^{-y} }{e^y + e^{-y} } = frac{ e^{2y} -1 }{e^{2y} +1 }.$$

Remember we want to solve for $y.$ Firstly solve for $u= e^{2y} $ first. Rearranging $ t= frac{u-1}{u+1} $ is simple, and hopefully you can do the rest.

$$tanh x=frac{e^{2x}-1}{e^{2x}+1} Rightarrow tanh frac{x}{2}=frac{e^{x}-1}{e^{x}+1} $$ In order to find $artanhfrac{x}{2}$ we have to solve following equation :

$$frac{x}{2}= frac {e^{2artanh frac{x}{2}}-1}{e^{2artanh frac{x}{2}}+1} Rightarrow x cdot e^{2artanh frac{x}{2}}+x=2e^{2artanh frac{x}{2}}-2 Rightarrow$$

$$x+2=(2-x) cdot e^{2artanh frac{x}{2}} Rightarrow e^{2artanh frac{x}{2}}=frac{2+x}{2-x} Rightarrow artanh frac{x}{2} =frac{1}{2} ln left(frac{2+x}{2-x}right) ; |x| < 2$$

Why is people not answering?

Look, it's quite intuitive so I'll just show you the step by step:

$y=tanh^{-1}x$

$tanh y=x$

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

$e^{y}-e^{-y}=xe^{y}+xe^{-y}$

$e^{2y}-1=xe^{2y}+x$

$e^{2y}(1-x)=1+x$

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

$y=frac{1}{2}ln(frac{1+x}{1-x})$