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Math Help - Please give me some hints:

  1. #1
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    Please give me some hints:

    I need to find the integral of:

    (2*x^(1/2))/(b+x^(1/2)), where b is a constant real number.

    thanks a lot,
    T
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Tuugii View Post
    I need to find the integral of:

    (2*x^(1/2))/(b+x^(1/2)), where b is a constant real number.

    thanks a lot,
    T
    there may be an easier way i'm not seeing, but it's 3am, cut me some slack. factor out the x^{1/2} from the denominator, then do a substitution of u = 2x^{1/2}. simplify as much as possible, then do another substitution of t = 2b + u. the resulting integral should be nice
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  3. #3
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    Substitution

    topsquark, if we let y = b + \sqrt{x}, then we have
    \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}
    dx = 2\sqrt{x} dy = 2(y - b) dy, right?

    (We could also write  y = b + \sqrt{x} \implies x = (y - b)^2 , so \dfrac{dx}{dy} = 2(y - b), and again, dx = 2(y - b) dy)

    Thus we have
    \int \frac{2\sqrt{x}}{b + \sqrt{x}}~dx = \int \frac{2(y - b)}{b + (y - b)} \cdot 2(y - b)~dy
    = 4 \int \frac{(y - b)^2}{y}~dy = 4 \int (y - 2b + \frac{b^2}{y})~dy

    For a noticeably different result.

    --Kevin C.
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  4. #4
    GAMMA Mathematics
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  5. #5
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    thanks

    thanks a lot guys!
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TwistedOne151 View Post
    topsquark, if we let y = b + \sqrt{x}, then we have
    \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}
    dx = 2\sqrt{x} dy = 2(y - b) dy, right?

    (We could also write  y = b + \sqrt{x} \implies x = (y - b)^2 , so \dfrac{dx}{dy} = 2(y - b), and again, dx = 2(y - b) dy)

    Thus we have
    \int \frac{2\sqrt{x}}{b + \sqrt{x}}~dx = \int \frac{2(y - b)}{b + (y - b)} \cdot 2(y - b)~dy
    = 4 \int \frac{(y - b)^2}{y}~dy = 4 \int (y - 2b + \frac{b^2}{y})~dy

    For a noticeably different result.

    --Kevin C.
    Yup. That's what I get for writing things down in the margin of my paper! Thanks for catching that.

    -Dan
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