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Math Help - Need helping solving an integral

  1. #1
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    Need helping solving an integral

    Hello, I can't manage to get anything out of this integral:

    \int^4_2 \sqrt{1+(x-\frac{1}{4x})^2}

    I've tried expanding and substituting, as well as factoring out different powers of x's and then substituting, but it never works.

    According to my calculus book, it's possible to do it (I'm doing arc length, so you get an abundance of integrals that don't work, but apparently, this one does...somehow)

    Ideas?

    Note: the original problem gave the equation y=\frac{x^2}{2}-\frac{ln}{4}, 2\leq x \leq 4, where it asked me to graph the curve and find its length using the arc length formula.
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  2. #2
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    Quote Originally Posted by RobLikesBrunch View Post
    Hello, I can't manage to get anything out of this integral:

    \int^4_2 \sqrt{1+(x-\frac{1}{4x})^2}

    I've tried expanding and substituting, as well as factoring out different powers of x's and then substituting, but it never works.

    According to my calculus book, it's possible to do it (I'm doing arc length, so you get an abundance of integrals that don't work, but apparently, this one does...somehow)

    Ideas?

    Note: the original problem gave the equation y=\frac{x^2}{2}-\frac{ln}{4}, 2\leq x \leq 4, where it asked me to graph the curve and find its length using the arc length formula.
    1+ \left ( x - \frac{1}{4x} \right)^2 = \left ( x + \frac{1}{4x} \right)^2.
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  3. #3
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    Well that was certainly elegant.

    I'm not entirely sure if why that works, though. How come you can simply change the sign by incorporating the 1?

    Anyway-- the integral works now. Woopie!
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  4. #4
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    Quote Originally Posted by RobLikesBrunch View Post
    Well that was certainly elegant.

    I'm not entirely sure if why that works, though. How come you can simply change the sign by incorporating the 1?
    a bit of algebra ...

    1 + \left(x - \frac{1}{4x}\right)^2

    1 + \left(x^2 - \frac{1}{2} + \frac{1}{16x^2}\right)

    x^2 + \frac{1}{2} + \frac{1}{16x^2}

    \left(x + \frac{1}{4x}\right)^2
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