How do you integrate int (x-3x^2)/((x-7)(x-5)(x+4)) ∫x−3x2(x−7)(x−5)(x+4) using partial fractions?
1 Answer
= -70/11 ln abs(x-7)+35/9 ln abs(x-5)-52/99 ln abs(x+4) + C=−7011ln|x−7|+359ln|x−5|−5299ln|x+4|+C
Explanation:
(x-3x^2)/((x-7)(x-5)(x+4)) = A/(x-7)+B/(x-5)+C/(x+4)x−3x2(x−7)(x−5)(x+4)=Ax−7+Bx−5+Cx+4
Using Heaviside's cover up method, we find:
A = ((7)-3(7)^2)/(((7)-5)((7)+4)) = (7-147)/(2*11) = -140/22 = -70/11A=(7)−3(7)2((7)−5)((7)+4)=7−1472⋅11=−14022=−7011
B = ((5)-3(5)^2)/(((5)-7)((5)+4)) = (5-75)/((-2)*9) = (-70)/(-18) = 35/9B=(5)−3(5)2((5)−7)((5)+4)=5−75(−2)⋅9=−70−18=359
C = ((-4)-3(-4)^2)/(((-4)-7)((-4)-5)) = (-4-48)/((-11)(-9)) = -52/99C=(−4)−3(−4)2((−4)−7)((−4)−5)=−4−48(−11)(−9)=−5299
So:
int (x-3x^2)/((x-7)(x-5)(x+4)) dx∫x−3x2(x−7)(x−5)(x+4)dx
= int -70/(11(x-7))+35/(9(x-5))-52/(99(x+4)) dx=∫−7011(x−7)+359(x−5)−5299(x+4)dx
= -70/11 ln abs(x-7)+35/9 ln abs(x-5)-52/99 ln abs(x+4) + C=−7011ln|x−7|+359ln|x−5|−5299ln|x+4|+C
(Note: The
