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Math Help - intergrating of rooted function

  1. #1
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    intergrating of rooted function

    Need help integrating \frac{\sqrt(x) - 4}{\sqrt(x) - 2}
    Last edited by foreg0ne; May 1st 2009 at 04:07 AM.
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  2. #2
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    Quote Originally Posted by foreg0ne View Post
    Need help deriving \frac{\sqrt(x) - 4}{\sqrt(x) - 2}
    Hang on.

    Are you finding an integral? A derivative? Or something else?

    Integrating is not the same as deriving and neither are the same as simplifying.

    Whatever the case, most likely the easiest way to solve your problem will be to simplify it first. To do this, multiply top and bottom by the conjugate of the bottom.


    \frac{\sqrt{x} - 4}{\sqrt{x} -2} = \frac{\sqrt{x} - 4}{\sqrt{x} -2} \times \frac{\sqrt{x} + 2}{\sqrt{x} + 2}

     = \frac{x - 2\sqrt{x} - 8}{x - 4}.


    Does this help?
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  3. #3
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    oops im sorry, i meant integrate (changed the title but forgot the text)
    -
    rationalizing makes sense...
    can i solve the rest by make .../(x-4) = ...(x-4)^-1 and integrate from there?
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  4. #4
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    Hello,

    I give you the answer, try to find the method :

    \int \frac{\sqrt{x}-4}{\sqrt{x}-2}=x-4\sqrt{x}-8\ln(\sqrt{x}-2)+C

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  5. #5
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    hmm this is what i can think of:

    \int \frac{\sqrt{x}-4}{\sqrt{x}-2} = u/v<br />
= \int ((v^2) ?? -4*2(v'))  dx

    i dont know how the root(x) comes in
    Last edited by foreg0ne; May 1st 2009 at 03:04 AM. Reason: forgot to put the 'dx' part
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  6. #6
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    Hi

    \int \frac{\sqrt{x}-4}{\sqrt{x}-2}\:dx=\int \left(1-2\frac{1}{\sqrt{x}-2}\right)\:dx = x-2\int \frac{dx}{\sqrt{x}-2}

    Set u = \sqrt x to calculate the right-hand integral
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  7. #7
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    I would be inclined to make the substitution u= \sqrt{x}= x^{1/2} immediately so that du= (1/2)x^{-1/2}dx= (1/2)(1/u)dx and 2u du= dx. The integral becomes \int\frac{\sqrt{x}-4}{\sqrt{x}- 2}dx= 2\int u\frac{u-4}{u-2}du
    Last edited by HallsofIvy; May 1st 2009 at 05:29 AM.
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  8. #8
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    are u guys leaving out steps or anything, because i dont follow the progression of ur working out
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  9. #9
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    Quote Originally Posted by HallsofIvy View Post
    I would be inclined to make the substitution u= \sqrt{x}= x^{1/2} so that du= (1/2)x^{-1/2}dx= (1/2)(1/u)dx and 2u du= dx. The integral becomes \int\frac{\sqrt{x}-4}{\sqrt{x}- 2}dx= 2\int u\frac{u-4}{u-2}du
    2\int \frac{u^2- 4u}{u-2}du= 2\int\left(u- 2- \frac{4}{u-2}\right)du
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