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Integrating Factor Differential Equation

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data



    3. The attempt at a solution

    so i split it into

    = ∫[itex]\frac{2x-1}{x^2-x}[/itex]dx - ∫[itex]\frac{x+1}{x^2-x}[/itex]dx

    = ln(x2-x) - ∫[itex]\frac{x}{x^2-x}[/itex] - ∫(x2-x)-1

    = ln(x2-x) - ln(x-1) - ∫(x2-x)-1

    ok. having problems working out ∫(x2-x)-1dx
    tried many ways but i keep ending up with the original integral.

    u=x-1 --> du=-x-2
    dv= (x-1)-1dx --> v=ln(x-1)

    gives me ([itex]\frac{1}{x}[/itex])ln(x-1) + ∫[itex]\frac{ln(x-1)}{x^2}[/itex]dx

    when i work this out


    u=ln(x-1) --> du=[itex]\frac{1}{x-1}[/itex]
    dv=x-2dx --> v=-x-1

    i get

    ∫[itex]\frac{ln(x-1)}{x^2}[/itex]dx = [itex]\frac{-ln(x-1)}{x}[/itex] + ∫(x2-x)-1

    which is the same integral and above and i get no solution.
    Need help......
  2. jcsd
  3. Feb 28, 2012 #2


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    Use partial fraction decomposition on [itex]\displaystyle\frac{x-2}{x(x-1)}\,.[/itex]
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