March 5th, 2008, 08:02 PM  #1 
Newbie Joined: Mar 2008 Posts: 5 Thanks: 0  partial fraction integral
doing partial fraction integrals: 4x5/(x^33x^2) 
March 6th, 2008, 12:49 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
One fraction is 7/(9(x  3)) by the cover up rule. Subtract that from the original function, cancel x  3 from numerator and denominator of the result, and the rest is easy.

March 6th, 2008, 03:03 PM  #3 
Member Joined: Feb 2008 Posts: 89 Thanks: 0 
Greetings: Do you mean (4x5)/(x^33x^2)? i.e., (4x5) / [(x3)x^2]? If so, let (4x5) / [(x3)x^2] = A/(x3) + B/x + C/x^2 ==> 4x5 = Ax^2 + Bx(x3) + C(x3) ==> 4x5 = (A+B)x^2 + (C3B)x  3C. Because 4x5 = 0x^2 + 4x  5, we equate the coefficients above as follows: A+B=0, C3B=4 and 3C=5. Finally, solving leaves A=7/9, B=7/9 and C=5/3. In conclusion, (4x5)/[x^33x^2) = 7/[9(x3)]  7/(9x) + 5/(3x^2) Regards, Rich B. 

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