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March 5th, 2008, 08:02 PM   #1
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partial fraction integral

doing partial fraction integrals:
4x-5/(x^3-3x^2)
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March 6th, 2008, 12:49 PM   #2
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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.
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March 6th, 2008, 03:03 PM   #3
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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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