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Partial Fraction Integration

  1. May 6, 2007 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution
    I know I should be using partial fractions for this problem but the x^2 in the denom. screws me up. I think the partial fraction should be:

    (Ax+B)/(x^2+4) + C/(3-x)





    So now Im completely lost assuming im even on the right track.:bugeye:

    Thanks Alot!!!
  2. jcsd
  3. May 6, 2007 #2


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    This makes no sense. Your first steps aren't actual equations, which is confusing you.

    (Ax+B)/(x^2+4) + C/(3-x) = (3x+4)/(x^2+4)(3-x)

    Multiplying by the two denomiator parts gives

    (Ax+B)(3-x) + c(x^2+4) = 3x+4

    This is for all x, so plug in clever values (like x=3), to solve for a, b and c
  4. May 6, 2007 #3


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    I'm not sure what you're doing! You have [tex]\frac{3x+4}{(x^2+4)(3-x)}=\frac{Ax+B}{x^2+4}+\frac{C}{3-x}[/tex]. Now multiply through by the denominator of the LHS to yield [tex]3x+4=(Ax+B)(3-x)+C(x^2+4)[/tex]. Now you need to find three equations to obtain A B and C.
  5. May 6, 2007 #4
    Ok it makes more sense now, this partial fraction stuff drives me nuts!:yuck:

    Thanks guys!
  6. May 7, 2007 #5
    Ok so I just worked on this problem again.... It was driving me nuts! This is what I came out with,

    So for a final answer I came up with,
    (Ln(x^2+4))/2+Ln(3-x)+ C

    Is that right??


  7. May 7, 2007 #6


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    Partial fraction decomposition right. Integration wrong. The second term should have a negative sign. The answer should be [tex]\frac{1}{2}\ln{(x^2+4)} - \ln{|3-x|} + C[/tex]. Don't forget the modulus sign on the second term (it's optional for the first since the square expression is nonnegative for real x).
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