Prove that #tanh^-1((x^2-1)/(x^2+1))=lnx# ?

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Answer 1 · 2018-07-14T16:20:05

Please see the proof below.

Let

#y=tanh^-1((x^2-1)/(x^2+1))#

#=>#, #((x^2-1)/(x^2+1))=tanhy#

#=>#, #((x^2-1)/(x^2+1))=(sinhy)/(coshy)#

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

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

#=(e^(2y)-1)/(e^(2y)+1)#

Cross multiply

#(e^(2y)+1)(x^2-1)=(e^(2y)-1)(x^2+1)#

Developing both sides

#x^2e^(2y)-e^(2y)+x^2-1=x^2e^(2y)+e^(2y)-x^2-1#

#2e^(2y)=2x^2#

#e^(2y)=x^2#

Taking log on both sides

#ln(e^(2y))=ln(x^2)#

#2y=2lnx#

#y=lnx#

Finally

#tanh^-1((x^2-1)/(x^2+1))=lnx#

#QED#