Question #71b42

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Answer 1 · 2017-10-12T15:22:14

The solution is #y=1/2ln((1+x)/(1-x))#

By definition

#tanhx=sinhx/coshx#

#coshx=(e^x+e^-x)/2#

#sinhx=(e^x-e^-x)/2#

If #y=tanh^-1x#

Then,

#x=tanhy#

#=sinhy/coshy#

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

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

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

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

So,

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

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

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

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

Taking natural logs

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

#2y=ln((1+x)/(1-x))#

#y=1/2ln((1+x)/(1-x))#

Conditions .

#AA x < |x|#