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Math Help - another arc length question

  1. #1
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    another arc length question

    Find the length of the curve, L.

    not exactly sure how to integrate this one after the derivative is found.. which is very long and complicated..
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  2. #2
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    Quote Originally Posted by khood View Post
    Find the length of the curve, L.

    not exactly sure how to integrate this one after the derivative is found.. which is very long and complicated..
    y=\ln \left( \frac{e^x+1}{e^x-1} \right)=\ln(e^x+1)-\ln(e^x-1)

    y'=\frac{e^x}{e^x+1}-\frac{e^x}{e^x-1}=\frac{e^{x}(e^x-1)-e^{x}(e^x+1)}{(e^{x}+1)(e^{x}-1)}=\frac{-2e^x}{(e^x+1)(e^x-1)}

     <br />
\int \sqrt{1+\left( \frac{-2e^x}{(e^x+1)(e^x-1)}\right)^2}dx=\int \sqrt{\frac{[(e^x+1)(e^x-1)]^2+4e^{2x}}{[(e^x+1)(e^x-1)]^2}}dx<br />

    \int \sqrt{\frac{[e^{2x}-1]^2+4e^{2x}}{[(e^x+1)(e^x-1)]^2}}dx=\int \sqrt{\frac{[e^{4x}-2e^{2x}+1+4e^{2x}}{[(e^x+1)(e^x-1)]^2}}dx

    \int \sqrt{\frac{e^{4x}+2e^{2x}+1}{[(e^x+1)(e^x-1)]^2}}dx = \int \sqrt{ \frac{[e^{2x}+1]^2}{[(e^x+1)(e^x-1)]^2}}dx

    \int \frac{e^{2x}+1}{(e^{x}+1)(e^{x}-1)}dx =\int \frac{e^{2x}+1}{e^{2x}-1}dx

    from here let u = e^{2x} \implies du=2e^{2x}dx \iff du=2udx \iff \frac{du}{2u}=dx

    \int \frac{u+1}{u-1}\frac{du}{2u}=\int\frac{u+1}{u(u-1)}du this can be integrated using partial fractions. Good luck.
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