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Need help finding some unknown constants

  1. Sep 23, 2011 #1
    1. The problem statement, all variables and given/known data

    I have an integral that I'm trying to split into partial fractions, and I've gotten to an equation but I'm not sure how to solve this one.

    2. Relevant equations



    3. The attempt at a solution

    93x+2 = Ax^2 + Bx + A - 4B

    I have no idea, because I usually don't have a x^2 term to deal with. I assumed that A should be 0 because of this, but that requires B to equal 93, and for A - 4B to equal 2. So, that's not the right direction.
     
  2. jcsd
  3. Sep 23, 2011 #2
    Yes A = 0

    and A - 4B = 2

    You know what A is...
     
  4. Sep 23, 2011 #3
    How does that work? That means B is - 1/2, and

    93x + 2 = -1/2x + 2

    Is not true.
     
  5. Sep 23, 2011 #4

    SammyS

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    And this requires B = -1/2 , but that contradicts B = 93.

    So, there appears to be something wrong with the way you obtained your equation: 93x+2 = Ax2 + Bx + A - 4B.

    Show us how you came up with that so we can help you.
     
  6. Sep 23, 2011 #5
    Okay

    [itex]\int \frac{93x+2}{x^{3}-4x^{2}+x-4}dx[/itex]

    I factored this by finding the factor of 4 that yields a 0 when plugged into the denominator, which is 4 itself. So I concluded one term is (x-4), and found the other through polynomial long division.

    [itex]\int \frac{93x+2}{(x-4)(x^{2}+1)}dx[/itex]

    So I then attempted to split this into partial fractions.


    [itex]\int \frac{A}{(x-4)}+ \frac{B}{(x^{2}+1)}dx[/itex]

    So then

    93x + 2 = Ax^2 + A + Bx - 4B

    What's wrong with the equation?
     
  7. Sep 23, 2011 #6

    SammyS

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    That needs to be [itex]\displaystyle \frac{A}{x-4}+ \frac{Bx + C}{x^{2}+1}[/itex]
     
  8. Sep 23, 2011 #7
    Oh, ok. Why is that? Because I have a variable in my numerator?
     
  9. Sep 23, 2011 #8

    SammyS

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    The denominator x2 + 1 is quadratic in x, so the most general numerator is linear in x.
     
  10. Sep 23, 2011 #9
    When working it I get:

    A = 23.25
    B = - A
    C = 5.3125

    Agree?
     
  11. Sep 23, 2011 #10

    SammyS

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    No, although, B = -A is correct.

    Did you get

    93x + 2 = A(x2 + 1) +(Bx +C)(x - 4) ?

    Hint: Set x = 4 to find A.
     
  12. Sep 25, 2011 #11
    I get

    x^2(A+B) + x(C-4B) - 4C + A = 93x + 2

    From there, I said that A+B = 0, C-4B = 93, and -4C + A = 2 and I am getting it wrong every time.
     
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