How do you integrate int (x-3x^2)/((x-7)(x-5)(x+4)) using partial fractions?
1 Answer
Sep 4, 2016
= -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
