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Partial fraction decomposition

  1. Apr 3, 2005 #1
    Use partial fraction decompostion to find:
    [tex]\int_{a}^{b} \frac{2x-1}{x^2(3x+1)(x^2 + 1)}[/tex]

    is this partial fraction set up correct?

    [tex]\frac{A}{3x +1} + \frac{Bx + C}{x^2 +1} + \frac{Dx + E}{x^2} = 2x - 1[/tex]

    If this is correct i can solve the integral.
     
  2. jcsd
  3. Apr 3, 2005 #2
    Almost. It's

    [tex]\frac{A}{3x +1} + \frac{Bx + C}{x^2 +1} + \frac{Dx + E}{x^2} = \frac{2x - 1}{x^2(3x+1)(x^2+1)}[/tex]

    I'm sure you meant that anyways, though...
     
  4. Apr 3, 2005 #3
    yeah that's what i meant. thanks
     
  5. Apr 3, 2005 #4
    this problem is harder than i thought are the following answers correct:

    A = 27/10
    B = -1
    C = 1/10
    D = -7/10
    E = 1
     
  6. Apr 3, 2005 #5
    I get

    [tex]A = -\ \frac{27}{2}, \; B = C = - \ \frac{1}{2}, \; D = 5, \; E = -1.[/tex]
     
  7. Apr 3, 2005 #6
    i don't see how Data. i believe you but i can't reproduce those numbers.
    how did you do it?
     
  8. Apr 3, 2005 #7
    my first number A = 27/10 is wrong so that makes every thing that follows wrong.
     
  9. Apr 3, 2005 #8
    OMG!!! i'm using 2x +1 instead of the correct 2x-1
     
  10. Apr 3, 2005 #9
    First multiply each side through by all three factors in the denominator on the right to get

    [tex]Ax^2(x^2+1) \ + \ (Bx+C)(3x+1)x^2 \ + \ (Dx+E)(x^2+1)(3x+1) = 2x-1[/tex]

    now sub [itex]x=0[/itex] to immediately give [itex]E=-1[/itex]. Then sub [itex]x=-1/3[/itex] to give

    [tex]A\left(-\frac{1}{3}\right)^2\left(\left(-\frac{1}{3}\right)^2+1\right) = -\frac{5}{3} \Longrightarrow A = -\frac{27}{2}[/tex]

    After that, look at the coefficient of [itex]x[/itex] on both sides. On the left it is [itex]D+3E[/itex] and on the right [itex]2[/itex], so

    [tex]D+3E = D - 3 = 2 \Longrightarrow D = 5.[/tex]

    Now look at the coefficient of [itex]x^2[/itex] on both sides. On the left it is [itex]A + C + 3D + E[/itex] and on the right [itex]0[/itex] so

    [tex] A + C + 3D + E = -\frac{27}{2} + C + 15 - 1 = 0 \Longrightarrow C = -\frac{1}{2}.[/tex]

    Finally look at the coefficient of [itex]x^3[/itex] on both sides. On the left it is [itex]B+3C+D+3E[/itex] and on the right it is [itex]0[/itex] so we get

    [tex]B + 3C + D + 3E = B - \frac{3}{2} + 5 - 3 = 0 \Longrightarrow B = -\frac{1}{2}[/tex]

    so overall we have found

    [tex] E = -1, \; A = -\frac{27}{2}, \; D = 5, \; C = B = -\frac{1}{2}[/tex]

    as I said :smile:
     
  11. Apr 3, 2005 #10
    I've been making dumb mistakes all day. thanks for taking the time to latex your replies , and thanks for all the help.
     
  12. Apr 3, 2005 #11
    No problem :smile:

    I latex everything that I can anyways, better to get used to it now!
     
  13. Apr 4, 2005 #12

    dextercioby

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    [tex]\frac{2x-1}{x^2\left( 3x+1\right) \left( x^2+1\right) }=\allowbreak -\frac 1{x^2}+\frac 5x-\frac{27}{2\left( 3x+1\right) }-\frac 12\frac{x+1}{x^2+1}[/tex]

    Daniel.

    P.S.Just trying my Maple :uhh: :tongue2: ...
     
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