`tanh^-1 x = 1/2 ln((1+x)/(1-x)) , -1

Given to prove

`tanh^(-1) x =1/2 ln((1+x)/(1-x))`

so let

`tanh^(-1) x =y`

=> `x= tanh(y)`

       `x =(e^y - e^-y)/(e^y + e^-y)`

=> `(e^y + e^-y)*x = (e^y - e^-y )`

=> `xe^y + xe^-y = e^y - e^-y`

=> `(xe^(2y)+x)/e^y = (e^2y -1)/e^y`

=> `(xe^(2y)+x)= (e^(2y) -1)`

=>`(xe^(2y)+x)-e^(2y) +1=0`

=>`e^(2y)(x-1)+x+1=0`

=>`(x-1)(e^(2y)) =-(x+1)`

=>`e^(2y) = -(x+1)/(x-1)`

=>` e^(2y) = (1+x)/(1-x)`

=>` e^(2y)=(1+x)/(1-x)`

=> `e^(2y) = ((1+x)/(1-x))`

=>`2y=ln (((1+x)/(1-x)))`

=>`y=1/2 ln (((1+x)/(1-x)))`

so,

`tanh^(-1) x =1/2 ln((1+x)/(1-x))`