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Math experts please come

  1. Feb 23, 2007 #1
    Math experts please come!!

    b]1. The problem statement, all variables and given/known data[/b]

    Calculus, integration

    2. Relevant equations
    [​IMG]


    3. The attempt at a solution

    used method:changing variable, by part, trigo etc but no avail.:confused: Please help me solve it, thanks.
     
    Last edited: Feb 23, 2007
  2. jcsd
  3. Feb 23, 2007 #2
    SD, Your kidding, right? Didn't your high school math teacher make you do a "time out" whenever misused the square root that way?
     
    Last edited: Feb 23, 2007
  4. Feb 23, 2007 #3
    Rewrite the argument x^2-x by "completing the square" . "u" substitution follows.
     
  5. Feb 23, 2007 #4

    Did you try a U substition with U = X^2 - X and dU = 2x - 1 dx?
     
  6. Feb 23, 2007 #5

    VietDao29

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    When you meet the problem asking you to integrate something like:
    [tex]\int \frac{dx}{(ax + b) ^ {m} \sqrt{c x ^ 2 + dx + e}}[/tex], what you should do is to let:
    [tex]ax + b = \frac{1}{t}[/tex]
    --------------------
    So, here, we go:
    [tex]\int \frac{dx}{(2x + 1) \sqrt{x ^ 2 - x}}[/tex]
    For [tex]\sqrt{x ^ 2 - x}[/tex] to be valid, x can only be in the interval [tex]] - \infty ; 0 ] \unity [1 ; + \infty [ [/tex]
    Let
    [tex]2x + 1 = \frac{1}{t} \quad \mbox{or} \quad x = \frac{1}{2} \left( \frac{1}{t} - 1 \right)[/tex]
    [tex]\Rightarrow 2dx = -\frac{dt}{t ^ 2}[/tex]
    The integral becomes:
    [tex]- \int \frac{\frac{dt}{2 t ^ 2}}{\frac{1}{t} \sqrt{ \left[ \frac{1}{2} \left( \frac{1}{t} - 1 \right) \right] ^ 2 - \frac{1}{2} \left( \frac{1}{t} - 1 \right)}}[/tex]

    [tex]= - \int \frac{dt}{2t\sqrt{ \left[ \frac{1}{4} \left( \frac{1}{t ^ 2} - \frac{2}{t} + 1 \right) \right] - \left( \frac{1}{2t} - \frac{1}{2} \right)}}[/tex]

    [tex]= - \int \frac{dt}{2t\sqrt{\frac{1}{4 t ^ 2} - \frac{1}{t} + \frac{3}{4}}}[/tex]

    [tex]= - \int \frac{dt}{t \sqrt{\frac{1}{t ^ 2} - \frac{4}{t} + 3}}[/tex]

    [tex]= \pm \int \frac{dt}{\sqrt{1 - 4t + 3 t ^ 2}}[/tex]
    Can you go from here? :)
     
    Last edited: Feb 23, 2007
  7. Feb 23, 2007 #6
    Forget what I had written here
     
    Last edited: Feb 23, 2007
  8. Feb 23, 2007 #7
    Won't work because the integrand has dx/(2x-1) and not (2x-1)dx.

    The best approach here is partial fraction decomp.

    Let 1=A(2x-1) + B(x^2-x)^1/2

    Set x=1 , A=1
    Set x=1/2 , B=2i

    Then integrate the new functions 1/(x^2-x)^1/2 and 2i/(2x-1)

    EDIT: The function 1/(x^2-x)^1/2 can be broken down further into |x^1/2|(x-1)^1/2 or into ((x-x^1/2)(x+x^1/2))^1/2, though the first of these two should look familiar.
     
    Last edited: Feb 23, 2007
  9. Feb 23, 2007 #8
    (x^2-x)^(1/2) = x - x^(1/2) is not true. You can check by multiplying the RHS with itself.
     
  10. Feb 23, 2007 #9

    arildno

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    Dearly Missed

    [tex]\sqrt{4^{2}-4}=\sqrt{16-4}=\sqrt{12}=2=4-\sqrt{4}[/tex]
    Fascinating..
     
  11. Feb 23, 2007 #10
    OK my bad. I was thinking of [tex]\sqrt{x^2} - \sqrt{x}[/tex]

    In my defence I had just finished a long and arduous day at work and wasn't really thinking:eek: sorry.

    I deleted it, still the shame lives on.... I'll go bury myself under a pile of soil and pretend I'm dead :smile:

    For my next trick I'll prove Fermat's last theorem in a new and exciting way whilst drunk :/
     
    Last edited: Feb 23, 2007
  12. Feb 23, 2007 #11
    going with my first reply

    Opps, I got fooled. Monkey Man's answer seemed so slick. I now vote for my own approach. It boils down to a constant times integral of
    1/(u sqrt(u^2 -a^2))


    "complete the square."

    I don't trust partial fractions since we don't have a rational function.
     
  13. Feb 23, 2007 #12
    Your kidding right? So 1/((2x-1)(x^2-x)^1/2) is not a rational function?
     
  14. Feb 23, 2007 #13
    Nope, It's not. Rational functions are ratios of polynomials.
     
  15. Feb 23, 2007 #14
    So 1/x is no longer rational?
     
  16. Feb 23, 2007 #15
    is too

    (no square root in that one)
     
  17. Feb 23, 2007 #16
    what does the square root matter?
     
  18. Feb 23, 2007 #17
    In that case, u=2x-1

    1/u(((u+1)2)^2-(u+1)/2)^1/2
     
  19. Feb 23, 2007 #18

    Hootenanny

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    Look up the definition of a polynomial.
     
  20. Feb 23, 2007 #19
    Thanks vietdao29. Ya i can go from there...I am a form 4 student studying integration on my own. Hope these can help me in form 5 and form 6.
     
  21. Feb 24, 2007 #20
    There is an easier way. I had the quickest way expect that I got diss'ed over partial fractions. You shouldn't have to rely on megaformulas.
     
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