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Find the exact length of the curve y=ln((1+e^(-x))/(1-e^(-x))) between x=1 and x=2
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\[Length = \int\limits_{a}^{b}\sqrt{1+f'(x)^{2}} dx\]
the derivative on wolfram is 2e^x/e^(2x)-1
i get f'(x) \[f'(x) = \frac{-2e^{-x}}{(1+e^{-x})(1-e^{-x})}\]
oh ok, well i didn't simplify it enough
\[\int\limits_{1}^{2}\sqrt{1+4e^(2x)/(e^(2x)-1)^2}\]
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i dont know how to integrate that
I haven't worked it out...but you should probably combine to make one term under the radical
what I wrote above is what you want to do
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