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Integrals of Rational Functions

  1. Mar 5, 2005 #1
    Integrals of Rational Functions....

    The integral of:....... (x-1)/x^4+6x^3+9x^2 , dx.........i factored out the bottom getting: x^2(x+3)(x+3).....so, my new integral is: (x-1)/x^2(x+3)^2...... now when i muiltlpy both sides by (x-1)/x^2(x+3)^2......i get............ x-1= A(x+3)^2 + Bx^2(x+3) + C x^2........for A i got -1/9....and C= -4/9......B iam not getting an answer for that....A,B,C are coefficients once finding them you plug them back into the second equation...that will give the final answer....need help.... :confused:
     
  2. jcsd
  3. Mar 6, 2005 #2
    You have to do it like this
    [tex]\frac{x-1}{x^{2}(x+3)^{2}}=\frac{A}{x}+\frac{Bx+C}{x^{2}}+\frac{D}{x+3}+\frac{Ex+F}{(x+3)^{2}}[/tex]
     
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