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 March 5th, 2008, 08:02 PM #1 Newbie   Joined: Mar 2008 Posts: 5 Thanks: 0 partial fraction integral doing partial fraction integrals: 4x-5/(x^3-3x^2)
 March 6th, 2008, 12:49 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2217 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 (4x-5)/(x^3-3x^2)? i.e., (4x-5) / [(x-3)x^2]? If so, let (4x-5) / [(x-3)x^2] = A/(x-3) + B/x + C/x^2 ==> 4x-5 = Ax^2 + Bx(x-3) + C(x-3) ==> 4x-5 = (A+B)x^2 + (C-3B)x - 3C. Because 4x-5 = 0x^2 + 4x - 5, we equate the coefficients above as follows: A+B=0, C-3B=4 and -3C=-5. Finally, solving leaves A=7/9, B=-7/9 and C=5/3. In conclusion, (4x-5)/[x^3-3x^2) = 7/[9(x-3)] - 7/(9x) + 5/(3x^2) Regards, Rich B.

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### 2/(4x^2 -1) as a partial fraction

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