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

  • Thread starter expscv
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  • #1
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having problem with intergrate :

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


thx
 
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Answers and Replies

  • #2
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btw how to use maths input thing?
 
  • #3
matt grime
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trig and latex, in that order.
 
  • #4
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huh? i m not sure wats exactly.
 
  • #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.
 
  • #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
 
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  • #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.
 
  • #8
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[tex] \int \frac{4}{x^2-2x-1} dx = \int \frac{4}{(x-1)^2-2}dx [/tex]?


great thx
 
  • #9
matt grime
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ah, apologies, it's hyperbolic trig, not trig.
 

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