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Homework Help: Hard integral

  1. Apr 22, 2005 #1
    i cant resolve this integral :confused:
    2x^3-2x^2+1/x^2-x-2?
     
  2. jcsd
  3. Apr 22, 2005 #2
    Is that

    [tex]\frac{2x^3-2x^2+1}{x^2-x-2}[/tex]

    ?

    If so, use polynomial longdivision and then partial fractions to transform the expression into something which can be integrated more easily.
     
  4. Apr 22, 2005 #3
    yeah, can you so kind to do some procedures.. cuz i did with partial fractions
    but the result is not correct =(
    thx much!
     
  5. Apr 22, 2005 #4

    Ba

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    Partial fractions don't work because the denominator is less than the numerator, use Muzza's idea and use long division.
     
  6. Apr 22, 2005 #5

    Hurkyl

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    Long division is part of the partial fractions algorithm!
     
  7. Apr 22, 2005 #6
    [tex]\int \frac{2x^3-2x^2+1}{x^2-x-2}[/tex]
    [tex]\frac{2x^3-2x^2+1}{x^2-x-2}=2x+\frac{2x+1}{2x^3-2x^2+1}[/tex]
    [tex]\int 2xdx + \int \frac{2x+1}{2x^3-2x^2+1}dx[/tex]
    now, use partial fractions on the second half. it shouldnt be too hard!
     
  8. Apr 22, 2005 #7

    dextercioby

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    The denominator is [itex](x-2)(x+1) [/itex],so it shouldn't be 2 difficult to get the simple fractions...



    Daniel.
     
  9. Apr 24, 2005 #8
    Did you mean

    [tex]\frac{2x^3-2x^2+1}{x^2-x-2}=2x+\frac{4x+1}{x^2-x-2}[/tex]?
     
  10. Apr 25, 2005 #9

    dextercioby

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    Yes,of course,yours is correct.He couldn't have changed the denominator.I mean he did,but it was wrong.

    Daniel.
     
  11. Jun 29, 2011 #10
    This is correct.

    Partial Fraction Decompisition for [itex]\frac{4x+1}{x^2-x-2}[/itex] is:


    [itex]\frac{4x+1}{(x+1)(x-2)}[/itex] = [itex]\frac{1}{x+1} + \frac{3}{x-2}[/itex]

    New Integral is:
    [itex]\int (2x + \frac{1}{x+1} + \frac{3}{x-2})dx[/itex]

    so,

    [itex]\int(\frac{2x^3-2x^2+1}{x^2-x-2})dx = x^2 + ln|x+1| + 3*ln|x-2| + C[/itex]
     
  12. Dec 22, 2013 #11
    x^2 + 3 Log[2 - x] + Log[1 + x]
     
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