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Partial fractions integral

  1. Feb 8, 2015 #1

    462chevelle

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    1. The problem statement, all variables and given/known data
    integral(0>1) of (x^2+x)/(x^2+x+1)dx

    2. Relevant equations
    Factor denominator, and set numerator with A,B,C, etc. multiply both sides by the common denominator.

    3. The attempt at a solution
    Since the denominator won't factor at all I don't really know where to start, I could rewrite it into 2 fractions or factor the numerator. Both useless though, I can't find an example in my book for this type of problem.

    Any hints on the first step?
    Thanks.
     
  2. jcsd
  3. Feb 8, 2015 #2

    PeroK

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    It's not partial fractions. The numerator and denominator are similar. Can you make use of that?
     
  4. Feb 8, 2015 #3

    462chevelle

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    Almost, as far as I can tell, if I take the denominator to be u I get du/(2x+1)=dx then I can factor a x out of the numerator and i get x(x-1)/(u(2x+1)). Unless I'm missing or forgetting something.
     
  5. Feb 8, 2015 #4

    462chevelle

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    Sorry, I'm bad at latex so I'm posting a picture of my work. I can't remember if that is allowed or not.
     

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  6. Feb 8, 2015 #5

    PeroK

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    How would you make the numerator the same as the denominator? Something very simple.
     
  7. Feb 8, 2015 #6

    462chevelle

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    You could add 2x+1 and that would be the derivative of the denominator. Then I would have to do that to the denominator. I'll see what I can figure out and post my work.
     
  8. Feb 8, 2015 #7

    PeroK

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    You're missing the obvious thing and going for more complicated things. What is the difference between the top and bottom?
     
  9. Feb 8, 2015 #8

    462chevelle

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    A constant and a sign.
     
  10. Feb 8, 2015 #9

    PeroK

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    Do you mean 1?
     
  11. Feb 8, 2015 #10

    462chevelle

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    Well, yes the constant would be 1. I just realized I made a typo on the first post, sorry. Its (x^2-x)/(x^2+x+1) the problem is correct in the picture I posted.
     
  12. Feb 8, 2015 #11

    PeroK

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    That makes the solution a lot easier you'll be glad to know!

    Let me do the original one and see if that helps:

    ##\frac{x^2 + x}{x^2+x+1} = 1 - \frac{1}{x^2+x+1}##
     
  13. Feb 8, 2015 #12

    HallsofIvy

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    Since numerator and denominator have the same degree, first do the division to get a polynomial plus a fraction with numerator of lower degree than the denominator. Then complete the square in the denominator.
     
  14. Feb 8, 2015 #13

    462chevelle

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    HMM, this seems to make it easy. I would have never thought of that otherwise.
     

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  15. Feb 8, 2015 #14

    462chevelle

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    I think I got it, thanks. I would have been beating my head against the wall all day otherwise.
     

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  16. Feb 8, 2015 #15

    Mark44

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    @462chevelle, since we saved you a whole day of beating your head against the wall, maybe you could devote some of that saved time to learning a bit of LaTeX? It's really not very hard. We have a brief summary here: https://www.physicsforums.com/help/latexhelp/

    Here are a few of the things you could have used in your problem. Note that I have omitted the # # pairs (without the space) at the beginning and end in my examples below. I did that so that the LaTeX script remains visible.

    Exponents:
    x^2
    e^{x + 1}

    Fractions:
    \frac{x^2 + x}{x^2 + x + 1}

    Integrals:
    \int 3x^2 dx

    \int_0^3 e^{x + 1} dx
     
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