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))`
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