How do you integrate #int(x)/((x-4)(x+2)(2x-1))# using partial fractions?
1 Answer
Jul 29, 2016
#int x/((x-4)(x+2)(2x-1)) dx#
#= 2/21 ln abs(x-4)-1/15 ln abs(x+2) - 1/35 ln abs(2x-1) + C#
Explanation:
#x/((x-4)(x+2)(2x-1)) = A/(x-4)+B/(x+2)+C/(2x-1)#
Use Heaviside's cover-up method to find:
#A = ((4))/(((4)+2)(2(4)-1)) = 4/((6)(7)) = 2/21#
#B = ((-2))/(((-2)-4)(2(-2)-1)) = (-2)/((-6)(-5)) = -1/15#
#C = ((1/2))/(((1/2)-4)((1/2)+2)) = 2/((1-8)(1+4)) = 2/((-7)(5)) = -2/35#
So:
#int x/((x-4)(x+2)(2x-1)) dx#
#=int 2/(21(x-4))-1/(15(x+2))-2/(35(2x-1)) dx#
#= 2/21 ln abs(x-4)-1/15 ln abs(x+2) - 1/35 ln abs(2x-1) + C#
