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Having problem with intergrate : _ _/ 4 / (x^2-2x-1)

  1. Mar 29, 2004 #1
    having problem with intergrate :

    ... _
    _/ 4 / (x^2-2x-1) dx


    thx
     
    Last edited: Mar 29, 2004
  2. jcsd
  3. Mar 29, 2004 #2
    btw how to use maths input thing?
     
  4. Mar 29, 2004 #3

    matt grime

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    trig and latex, in that order.
     
  5. Mar 29, 2004 #4
    huh? i m not sure wats exactly.
     
  6. Mar 29, 2004 #5

    matt grime

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    ok, to do the integral requires a trig subsitution, and to do the images, which is what I think you meant by "maths input thing" you need to use latex, which isnt' as odd as it might appear. there's a sticky thread somewhere explaining it, I beleieve if you go ot the general physics forum it's the first thread there. it takes a little memorizing but it's worth it.
     
  7. Mar 29, 2004 #6

    matt grime

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    [tex] \int\frac{1}{x^2-2x-1}dx[/tex]

    click on the image and it should show you the source code for it
     
    Last edited: Mar 29, 2004
  8. Mar 29, 2004 #7

    HallsofIvy

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    Getting back to the original problem, to integrate [itex] \int\frac{4}{x^2-2x-1}dx[/itex] try completing the square in the denominator: x2- 2x- 1= x2- 2x+ 1- 2= (x-1)2-2. Make the change of variable u= x-1 and the integral becomes [itex]\int\frac{4}{u^2-2}du[/itex]. Now the denominator factors as
    [itex]u^2-2= (u-\sqrt{2})(u+\sqrt{2})[/itex] and the integral can be done by partial fractions.
     
  9. Mar 29, 2004 #8
    [tex] \int \frac{4}{x^2-2x-1} dx = \int \frac{4}{(x-1)^2-2}dx [/tex]?


    great thx
     
  10. Mar 29, 2004 #9

    matt grime

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    ah, apologies, it's hyperbolic trig, not trig.
     
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