How do you integrate f(x)=(x)/((x+4)(x-2)(2x-2)) using partial fractions?
1 Answer
May 29, 2017
=-1/15 ln abs(x+4)+1/6 ln abs(x-2)-1/10 ln abs(x-1) + C
Explanation:
x/((x+4)(x-2)(2x-2)) = x/(2(x+4)(x-2)(x-1))
color(white)(x/((x+4)(x-2)(2x-2))) = A/(x+4)+B/(x-2)+C/(x-1)
Use Oliver Heaviside's cover up method to find
A = (color(blue)(-4))/(2((color(blue)(-4))-2)((color(blue)(-4))-1)) = (-4)/(2(-6)(-5)) = -1/15
B = color(blue)(2)/(2(color(blue)(2)+4)(color(blue)(2)-1)) = 2/(2(6)(1)) = 1/6
C = color(blue)(1)/(2(color(blue)(1)+4)(color(blue)(1)-2)) = 1/(2(5)(-1)) = -1/10
So:
int x/((x+4)(x-2)(2x-2)) dx
=int (-1/15(1/(x+4))+1/6(1/(x-2))-1/10(1/(x-1))) dx
=-1/15 ln abs(x+4)+1/6 ln abs(x-2)-1/10 ln abs(x-1) + C
