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Homework Help: Integration by partial fractions.

  1. May 26, 2010 #1
    1.[tex]\int[/tex]2x^2+x+9/(9x+1)(x^2+9) dx

    2. (A/9x+1) + [(Bx + C ) / (x^2 + 9)]

    I get the worst numbers when I solve the system. The question is from an old exam and calculators are not allowed. Am I doing something wrong or is there another way to integrate this?
     
  2. jcsd
  3. May 26, 2010 #2

    Mark44

    Staff: Mentor

    Are you sure of the numbers you are getting? IOW, did you check to make sure that your two separate fractions add up to the larger one?
     
  4. May 26, 2010 #3
    Yes, I used a solver to check my work because it did not seem right. I am fairly sure the fractions are broken up properly as well. =(
     
  5. May 26, 2010 #4

    Mark44

    Staff: Mentor

    Yes, your decomposition is fine. Out of curiosity, what did you get for A, B, and C?
     
  6. May 26, 2010 #5
    a=37/41 b=5/41 c = 7/41 I mean its not that outrageous I guess. I guess I just assume I do things wrong lol.
     
  7. May 26, 2010 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
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    Just to make sure, the integrand is

    [tex]\frac{2x^2+x+9}{(9x+1)(x^2+9)}[/tex]

    and not

    [tex]2x^2+x+\frac{9}{(9x+1)(x^2+9)}[/tex]

    right? In either case, your values for A, B, and C aren't correct. If the first version of the integrand is the right one, the coefficients come out even less pretty.
     
  8. May 26, 2010 #7
    Yes the first version is correct. -.- Do you have any suggestions?
     
  9. May 26, 2010 #8
    your partial fractions expansion is:

    [tex]
    \frac{A}{9 x + 1} + \frac{B \, x + C}{x^{2} + 9}
    [/tex]

    Multiplying both sides by [itex](9 x + 1)(x^{2} + 9)[/itex] and comparing similar terms, you will get 3 equations for 3 unknowns.
     
  10. May 26, 2010 #9
    I get the following system:

    [tex]
    \left\{
    \begin{array}{rcl}
    9 A + C & = & 9 \\

    B + 9 C & = & 1 \\

    A + 9 B & = & 2
    \end{array} \right.
    [/tex]
     
  11. May 26, 2010 #10

    vela

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    After you multiply by the denominator, I'd set x=-1/9, which allows you to solve for A, then use the equations Dickfore got to solve for B and C.
     
  12. May 26, 2010 #11
    Yeah.... It must be some type of cruel joke.
     
  13. May 26, 2010 #12
    The denominators for the solutions are equal to the number of days in a year.
     
  14. May 26, 2010 #13
    361/365 41/365 36/365
    It was on an old exam lol. I hope there's nothing like that on our exam =O.
     
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