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Math Help - calc2. find the arc length of a function

  1. #1
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    calc2. find the arc length of a function

    y=ln[(e^x +1)/(e^x -1)]
    interval [ln2,ln3]

    please help

    thanks
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  2. #2
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    Krizalid's Avatar
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    I assume you know the arc-lenght formula for this case.

    Show your work please, or may be an attempt.
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  3. #3
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    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
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  4. #4
    Math Engineering Student
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    Quote Originally Posted by tichua View Post

    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 \frac{{{e}^{2x}}+1}{{{e}^{2x}}-1}=\frac{{{e}^{x}}+{{e}^{-x}}}{{{e}^{x}}-{{e}^{-x}}} and by differentiating e^x-e^{-x} you'll get e^x+e^{-x} which is the numerator, so I think you know what's the proper substitution.
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