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 .

$AA x < |x|$