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Partial fraction decomposition for .

  1. Jul 3, 2011 #1
    Partial fraction decomposition for.....

    1. 3x-1 / x(x^2 +4)


    2. Relevant equations



    3. A/x + Bx + C/ x^2 +4

    after multiplying through by the denominator and my attempt at finding A,B,C i get this:

    3x-1/x(x^2+4) = - one fourth / x + one fourth + 3/ x^2 +4. I dont feel comfortable about this answer.
     
  2. jcsd
  3. Jul 3, 2011 #2
    Re: Partial fraction decomposition for.....

    The answer i get is:
    -(1/4)ln(x)-(1/8)ln(x^2+4)+(3/2)arctan(x/4)+c

    i think you made a mistake at Bx which is supp0sed to be B(2x)/(x^2+4).
     
  4. Jul 3, 2011 #3

    eumyang

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    Re: Partial fraction decomposition for.....

    @itachi: Please either learn LaTeX or use parentheses. I originally read this:

    as this:
    [tex]3x - \frac{1}{x(x^2 + 4)}[/tex]

    Then I originally read this:
    as this:
    [tex]\frac{A}{x} + Bx + \frac{C}{x^2 + 4}[/tex]

    And finally your answer:
    I thought was this:
    [tex]-\frac{1/4}{x} + \frac{1}{4} + \frac{3}{x^2} + 4[/tex]

    But it looks like you meant this as your answer:
    [tex]-\frac{1/4}{x} + \frac{\frac{1}{4} + 3}{x^2 + 4}[/tex]

    If so, you forgot an x next to the 1/4 in the 2nd fraction.

    @median27, I'm not sure how you got your answer.
     
  5. Jul 3, 2011 #4
    Re: Partial fraction decomposition for.....

    Disregard my post. I apply integration. I thought your query about partial fractions was under integral calculus. I've never encounter partial fraction decomposition in other subjects other than in integral so i assumed it as a query about integration. :D
     
  6. Jul 3, 2011 #5

    SammyS

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    Re: Partial fraction decomposition for.....


    Parentheses are important !

    Show the rest of your work (with correct grouping) so we can see what you've done.
     
  7. Jul 3, 2011 #6
    Re: Partial fraction decomposition for.....

    sorry about the mistake i made with the problem. yes the problem is (3x-1)/(x(x^2 +4)). From the equation i came up with A/x + Bx + C/ x^2 +4. After multiplying through the equation by the denominator (x(x^2+4)) i get A(x^2 +4) + (Bx +C)(x). That gives me Ax^2 + 4A + Bx^2 + Cx. After equating coefficients i come up with A + B=0, C=3, and 4A=-1. I solve for 4A which gives me A= -1/4 and plug this into A + B=0 which gives me B= 1/4. This leads me to my answer which is (- one fourth / x) + (one fourth x + 3/ x^2 +4). Is this correct???
     
  8. Jul 3, 2011 #7

    eumyang

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    Re: Partial fraction decomposition for.....

    Don't mean to be so picky, but you're still not applying enough parentheses. This looks like
    [tex]\frac{A}{x} + Bx + \frac{C}{x^2} + 4[/tex].
    Without LaTeX, you should have typed
    A/x + (Bx + C)/(x^2 +4)

    And your answer looks like
    [tex]-\frac{1/4}{x} + \frac{1}{4}x + \frac{3}{x^2} + 4[/tex]

    But yes, I am getting the same values for A, B, and C.


    EDIT: Wow, 500 posts already?
     
  9. Jul 6, 2011 #8

    Ray Vickson

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    Re: Partial fraction decomposition for.....

    Is A/x + Bx + C/x^2+4 supposed to be A/x + Bx + 4 + C/x^2, or is it A/x + Bx + C/(x^2+4)? You don't need to use LaTeX (although it would be better if you did), but you should use brackets. Why wouldn't you? It is easy and avoids confustion.

    RGV
     
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