How do you integrate #f(x)=(x)/((x+4)(x-2)(2x-2))# using partial fractions?

1 Answer
George C.
May 29, 2017

#int x/((x+4)(x-2)(2x-2)) dx#

#=-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#, #B# and #C#:

#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#

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