How do you integrate #f(x)=(x^2+x)/((x^2-1)(x-4)(x-9))# using partial fractions?
1 Answer
Jul 29, 2016
#int (x^2+x)/((x^2-1)(x-4)(x-9)) dx#
#= 1/24 ln abs(x-1) - 4/15 ln abs(x-4)+9/40 ln abs(x-9) + C#
Explanation:
#(x^2+x)/((x^2-1)(x-4)(x-9))#
#=(x^2+x)/((x-1)(x+1)(x-4)(x-9))#
#=A/(x-1)+B/(x+1)+C/(x-4)+D/(x-9)#
Use Heaviside's cover-up method to find:
#A = ((1)^2+(1))/(((1)+1)((1)-4)((1)-9)) = 2/((2)(-3)(-8)) = 1/24#
#B = ((-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)) = -4/15#
#D = ((9)^2+(9))/(((9)-1)((9)+1)((9)-4)) = 90/((8)(10)(5)) = 9/40#
So:
#int (x^2+x)/((x^2-1)(x-4)(x-9)) dx#
#= int 1/(24(x-1))-4/(15(x-4))+9/(40(x-9)) dx#
#= 1/24 ln abs(x-1) - 4/15 ln abs(x-4)+9/40 ln abs(x-9) + C#
