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

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))