y=ln[(e^x +1)/(e^x -1)]
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I assume you know the arc-lenght formula for this case.
Show your work please, or may be an attempt.
y'= (-2e^x)/(e^2x - 1)
(y'^2)= [4e^(2x)]/[(e^2x - 1)^2]
sqroot of [1+ (y'^2)]= (e^2x +1) / (e^2x -1)
then i got stuck. i have no idea how to take the integral of (e^2x +1) / (e^2x -1)
Originally Posted by tichua
i have no idea how to take the integral of (e^2x +1) / (e^2x -1) I'll assume your work is fine.
Now, to integrate that, just observe that and by differentiating you'll get which is the numerator, so I think you know what's the proper substitution.
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