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Math Help - Arc Length Help

  1. #1
    Newbie DaveDammit's Avatar
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    Arc Length Help

    Hey guys, I love my Calc. II class and this arc length integral has me kind of stumped. I put the integral in my TI-83 calculator and got an answer but I wanted to see if any of you thought there would be a way to integrate this:

    \int\sqrt{1 + \frac{4}{(2x+1)^2}} dx on the interval [\frac{1}{2},\sqrt{3} - \frac{1}{2}]

    The original problem was calculate the arc length of y = \ln{2x+1} on the interval [\frac{1}{2},\sqrt{3} - \frac{1}{2}]

    Any ideas? Try to force a perfect square trinomial in there? Thanks for any help!
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  2. #2
    Eater of Worlds
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    \left[\frac{d}{dx}(ln(2x)+1)\right]^{2}=\frac{1}{x^{2}}

    Arc length would be \int_{\frac{1}{2}}^{\sqrt{3}-\frac{1}{2}}\sqrt{1+\frac{1}{x^{2}}}dx
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  3. #3
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    Go to this website Wolfram|Alpha
    In the input window, type in this exact expression: integrate sqrt[1 +4/(2x+1)^2]dx from 1/2 to sqrt[3]-(1/2) (you can copy & paste) Click the equals bar at the right-hand end of the input window..
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  4. #4
    Newbie DaveDammit's Avatar
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    Quote Originally Posted by galactus View Post
    \left[\frac{d}{dx}(ln(2x)+1)\right]^{2}=\frac{1}{x^{2}}

    Arc length would be \int_{\frac{1}{2}}^{\sqrt{3}-\frac{1}{2}}\sqrt{1+\frac{1}{x^{2}}}dx
    Oops.. I put in the code wrong. It's not y = \ln(2x) + 1 it's y = \ln(2x+1)

    Thanks for the Wolfram/Alpha link, I'm going to check it out.
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  5. #5
    Newbie DaveDammit's Avatar
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    Ok so I put the integral into Wolfram Alfa and it gave me the same answer that my calculator gave me when I plugged it into that. Even with the answer that Wolfram Alfa gives me I'm still struggling with how to integrate the problem.

    Is there maybe a way I can use trigonometric substitution to get a\tan\theta into the \frac{4}{(2x+1)^2} part of the function?

    Thanks for any more help!
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  6. #6
    Eater of Worlds
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    \int\sqrt{1+\frac{4}{(2x+1)^{2}}}dx

    If we let x=\frac{u-1}{2}, \;\ dx=du

    then it whittles down to:

    \int\frac{\sqrt{u^{2}+4}}{u}du

    Now, it is easier to integrate?.

    You can try different routes including trig sub.
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