y=ln[(e^x +1)/(e^x -1)]

interval [ln2,ln3]

please help

thanks

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- Feb 3rd 2009, 07:51 PMtichuacalc2. find the arc length of a function
y=ln[(e^x +1)/(e^x -1)]

interval [ln2,ln3]

please help

thanks - Feb 3rd 2009, 07:58 PMKrizalid
I assume you know the arc-lenght formula for this case.

Show your work please, or may be an attempt. - Feb 3rd 2009, 08:04 PMtichua
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)

thanks - Feb 5th 2009, 03:01 PMKrizalid
I'll assume your work is fine.

Now, to integrate that, just observe that $\displaystyle \frac{{{e}^{2x}}+1}{{{e}^{2x}}-1}=\frac{{{e}^{x}}+{{e}^{-x}}}{{{e}^{x}}-{{e}^{-x}}}$ and by differentiating $\displaystyle e^x-e^{-x}$ you'll get $\displaystyle e^x+e^{-x}$ which is the numerator, so I think you know what's the proper substitution. :eek: