How do you integrate #int (x-3x^2)/((x-7)(x-5)(x+4)) # using partial fractions?

1 Answer
George C.
Sep 4, 2016

#int (x-3x^2)/((x-7)(x-5)(x+4)) dx#

#= -70/11 ln abs(x-7)+35/9 ln abs(x-5)-52/99 ln abs(x+4) + C#

Explanation:

#(x-3x^2)/((x-7)(x-5)(x+4)) = A/(x-7)+B/(x-5)+C/(x+4)#

Using Heaviside's cover up method, we find:

#A = ((7)-3(7)^2)/(((7)-5)((7)+4)) = (7-147)/(2*11) = -140/22 = -70/11#

#B = ((5)-3(5)^2)/(((5)-7)((5)+4)) = (5-75)/((-2)*9) = (-70)/(-18) = 35/9#

#C = ((-4)-3(-4)^2)/(((-4)-7)((-4)-5)) = (-4-48)/((-11)(-9)) = -52/99#

So:

#int (x-3x^2)/((x-7)(x-5)(x+4)) dx#

#= int -70/(11(x-7))+35/(9(x-5))-52/(99(x+4)) dx#

#= -70/11 ln abs(x-7)+35/9 ln abs(x-5)-52/99 ln abs(x+4) + C#

(Note: The #C# here stands for the integration constant, not the coefficient we calculated).

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