How do you integrate #int(x)/((x+4)(x+2)(2x-11))# using partial fractions?
1 Answer
Jul 28, 2016
#int x/((x+4)(x+2)(2x-11)) dx#
#= -2/19 ln(abs(x+4)) + 1/15 ln(abs(x+2)) + 11/285 ln(abs(2x-11)) + C#
Explanation:
#x/((x+4)(x+2)(2x-11)) = A/(x+4)+B/(x+2)+C/(2x-11)#
Use Heaviside's cover up method to find:
#A = (-4)/(((-4)+2)(2(-4)-11)) = (-4)/((-2)(-19)) = -2/19#
#B = (-2)/(((-2)+4)(2(-2)-11)) = (-2)/((2)(-15)) = 1/15#
#C = (11/2)/(((11/2)+4)((11/2)+2)) = 22/((11+8)(11+4)) = 22/285#
So:
#int x/((x+4)(x+2)(2x-11)) dx#
#= int (-2/(19(x+4))+1/(15(x+2))+22/(285(2x-11))) dx#
#= -2/19 ln(abs(x+4)) + 1/15 ln(abs(x+2)) + 11/285 ln(abs(2x-11)) + C#
Notice the coefficient
