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Partial Fractions Explanation

  1. Apr 10, 2014 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations

    After looking through this on Wiki, I'm a little confused as to how these partial fractions are multiplied out. Is there a rule or something for this?

    With simpler partials I can do it but this one is something else!

    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Apr 10, 2014 #2

    HallsofIvy

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    The denominator on the left is [tex](x- 1)^3(x^2+ 1)^2[/tex] and every denominator on the right is a factor of that, so multiplying both sides by it will remove all fractions:

    [tex]\frac{2x^6- 4x^5+ 5x^4- 3x^3+ x^2+ 3x}{(x- 1)^3(x^2+ 1)^2}((x-1)^3(x^2+ 1)^2)= 2x^6- 4x^5+ 5x^4- 3x^3+ x^2+ 3x[/tex]

    [tex]\frac{A}{x- 1}((x-1)^3(x^2+ 1)^2)= A(x- 1)^2(x^2+ 1)^2[/tex]
    [tex]\frac{B}{(x- 1)^2}((x-1)^3(x^2+ 1)^2)= B(x- 1)(x^2+ 1)^2[/tex]
    [tex]\frac{C}{(x- 1)^3}((x-1)^3(x^2+ 1)^2)= C(x^2+ 1)^2[/tex]
    [tex]\frac{Dx+ E}{x^2+ 1}((x-1)^3(x^2+ 1)^2)= (Dx+ E)(x- 1)^3(x^2+ 1)[/tex]
    [tex]\frac{Fx+ G}{(x^2+ 1)^2}((x-1)^3(x^2+ 1)^2)= (Fx+ G)(x- 1)^3[/tex]
     
  4. Apr 10, 2014 #3
    Oh dear, had a complete 'idiot moment'. Thanks dude.

    I've actually been googling trying to find an online differentiation calculator but I'm sure that some of them aren't giving the correct answers- do you know of any which are reliable?
     
  5. Apr 10, 2014 #4

    Ray Vickson

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    Wolfram Alpha.
     
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