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How do I integrate this?

  1. Nov 24, 2007 #1
    [tex]\int\frac{3x^4+4x^3+16x^2+20x+9}{x^5+2x^4+6x^3+12x^2+9x+18}dx[/tex]

    I tried using du/u for Lnu but I couldn't get it into that form. I don't have any idea what to do next, really. Can someone point me in the right direction?
     
  2. jcsd
  3. Nov 25, 2007 #2

    Ben Niehoff

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    You should try expanding it via partial fractions. There should be a section in your textbook about this.
     
  4. Nov 25, 2007 #3

    HallsofIvy

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    I would have sworn that denominator couldn't be factored, but it can!
     
  5. Nov 25, 2007 #4
    take u = (x^5+2x^4+6x^3+12x^2+9x+18)
    du = 5x^4+8x^3+18x^2+24x+9
    then take the numerator & cancel it out with du.
     
  6. Nov 25, 2007 #5
    What?

    I'm having trouble factoring it, and I'm not sure what Nachore means.
     
  7. Nov 25, 2007 #6
    Use Kramer on the denominator, you will find that -2 is a root.
    The remaining part can be factorized easily.
    Then use partial fractions to split the integral into more classical ones.

    This is not a particularly difficult integral, although a very tedious one to solve.
    You might say that this makes it a difficult one after all.

    Edit: It is not as tedious as I thought at first, the solution is ln(x+2)+ln(x^2+3)-2/(x^2+3)+C
     
    Last edited: Nov 25, 2007
  8. Nov 25, 2007 #7

    HallsofIvy

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    Yes, that's true. but since the numerator is nothing like that, it doesn't help at all!
     
  9. Nov 27, 2007 #8
    I get (x+2)(x^2+3)^2 as the bottom, factored out. Is this correct? I'm really not good at factoring.
     
  10. Nov 27, 2007 #9

    HallsofIvy

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    Yes, that's correct. Now use partial fractions.
     
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