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Partial fractions

  1. Sep 24, 2009 #1
    1/(x+5)^2 (x-1)
    B=(-1/6) C=(1/36)

    I can't find the value of A, what method do you use to find it?
     
  2. jcsd
  3. Sep 24, 2009 #2

    rl.bhat

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    b]1/(x+5)^2 (x-1)[/b] = A/(x+5) + B/(x + 5)^2 + C/(x - 1)
    Put x = 0.
     
  4. Sep 24, 2009 #3
    [tex]\frac{1}{(x+5)^{2}(x-1)}[/tex] will disintegrate into [tex]\frac{Ax+B}{(x+5)^{2}} + \frac{C}{x-1}[/tex]

    Compare coefficients on both sides to get

    A = -1/36
    B = -11/36
    C = 1/36
     
  5. Sep 24, 2009 #4
    Ok thank, now what about this problem?

    [tex]\frac{(x)^3}{(x+4)^{2}}[/tex] will disintegrate into [tex]x+4[/tex]=[tex]\frac{Ax+B}{(x+4)^{2}}[/tex]

    I'm not sure where to go from there, to get the values of A & B.
     
  6. Sep 25, 2009 #5
    [tex]\frac{(x)^3}{(x+4)^{2}}[/tex] will disintegrate into

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

    which you can expand using the [tex](a+b)^3[\tex\ standard formula and then its the same as the last one. Compare coefficients of powers of x on both sides to get A,B,C and so on.
     
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