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Integrate (1+x+x^2) / x(1+x^2) for x. Partial dractions?

  1. Jan 21, 2006 #1
    integre (1+x+x^2) / x(1+x^2)
    pls help....
    i used partial fraction and obtain the answer of ln x - 1/2 ln (1-x^2) + arctan x
    is it correct??
     
  2. jcsd
  3. Jan 21, 2006 #2

    HallsofIvy

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    That's not what I get. HOW did you get that? What was the partial fractions expansion?
     
  4. Jan 21, 2006 #3
    i got 1 +x +x^2 = A(1+x^2) + B(x^2) + Cx
    then let x =1 and -1 and again x=0

    is it correct??
     
  5. Jan 21, 2006 #4

    VietDao29

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    Looks good. So what are your A, B, and C?
    Remember that:
    [tex]\frac{1 + x + x ^ 2}{x(1 + x ^ 2)} = \frac{A}{x} + \frac{Bx + C}{1 + x ^ 2}[/tex]
    What does your final result look like?
     
  6. Jan 21, 2006 #5
    A=1 B=0 and C=1
     
  7. Jan 21, 2006 #6
    is it right
     
  8. Jan 21, 2006 #7

    VietDao29

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    Yup, it looks good. So:
    [tex]\int \frac{1 + x + x ^ 2}{x(1 + x ^ 2)} dx = \int \left( \frac{1}{x} + \frac{1}{1 + x ^ 2} \right) dx = ?[/tex]
    Can you integrate this? And what's your answer, then?
     
  9. Jan 21, 2006 #8
    oh.....i intgre wrongly on the above....is should be ln x + arctan x rite??
     
  10. Jan 21, 2006 #9

    VietDao29

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    Looks about right, but don't forget the constant of integration. :wink:
     
  11. Jan 21, 2006 #10

    Ouabache

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    Are they teaching integration in precalculus now?:confused:
     
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