How do you integrate f(x)=(x^2+x)/((x^2-1)(x-4)(x-9))f(x)=x2+x(x2−1)(x−4)(x−9) using partial fractions?
1 Answer
int (x^2+x)/((x^2-1)(x-4)(x-9)) dx∫x2+x(x2−1)(x−4)(x−9)dx
= 1/24 ln abs(x-1) - 4/15 ln abs(x-4)+9/40 ln abs(x-9) + C=124ln|x−1|−415ln|x−4|+940ln|x−9|+C
Explanation:
(x^2+x)/((x^2-1)(x-4)(x-9))x2+x(x2−1)(x−4)(x−9)
=(x^2+x)/((x-1)(x+1)(x-4)(x-9))=x2+x(x−1)(x+1)(x−4)(x−9)
=A/(x-1)+B/(x+1)+C/(x-4)+D/(x-9)=Ax−1+Bx+1+Cx−4+Dx−9
Use Heaviside's cover-up method to find:
A = ((1)^2+(1))/(((1)+1)((1)-4)((1)-9)) = 2/((2)(-3)(-8)) = 1/24A=(1)2+(1)((1)+1)((1)−4)((1)−9)=2(2)(−3)(−8)=124
B = ((-1)^2+(-1))/(((-1)-1)((-1)-4)((-1)-9)) = 0/((-2)(-5)(-10)) = 0B=(−1)2+(−1)((−1)−1)((−1)−4)((−1)−9)=0(−2)(−5)(−10)=0
C = ((4)^2+(4))/(((4)-1)((4)+1)((4)-9)) = 20/((3)(5)(-5)) = -4/15C=(4)2+(4)((4)−1)((4)+1)((4)−9)=20(3)(5)(−5)=−415
D = ((9)^2+(9))/(((9)-1)((9)+1)((9)-4)) = 90/((8)(10)(5)) = 9/40D=(9)2+(9)((9)−1)((9)+1)((9)−4)=90(8)(10)(5)=940
So:
int (x^2+x)/((x^2-1)(x-4)(x-9)) dx∫x2+x(x2−1)(x−4)(x−9)dx
= int 1/(24(x-1))-4/(15(x-4))+9/(40(x-9)) dx=∫124(x−1)−415(x−4)+940(x−9)dx
= 1/24 ln abs(x-1) - 4/15 ln abs(x-4)+9/40 ln abs(x-9) + C=124ln|x−1|−415ln|x−4|+940ln|x−9|+C
