How do you integrate #int (1-2x^2)/((x-8)(x+6)(x-7)) # using partial fractions?
1 Answer
Oct 12, 2016
#= -127/14 ln abs(x-8) -71/182ln abs(x+6)+97/13 ln abs(x-7) + C#
Explanation:
#(1-2x^2)/((x-8)(x+6)(x-7)) = a/(x-8)+b/(x+6)+c/(x-7)#
Use Heaviside's cover up method to find
#a = (1-2(color(blue)(8))^2)/(((color(blue)(8))+6)((color(blue)(8))-7)) = (1-128)/((14)(1)) = -127/14#
#b = (1-2(color(blue)(-6))^2)/(((color(blue)(-6))-8)((color(blue)(-6))-7)) = (1-72)/((-14)(-13)) = -71/182#
#c = (1-2(color(blue)(7))^2)/(((color(blue)(7))-8)((color(blue)(7))+6)) = (1-98)/((-1)(13)) = 97/13#
So:
#int (1-2x^2)/((x-8)(x+6)(x-7)) dx#
#= int -(127/14)(1/(x-8)) -(71/182)(1/(x+6))+(97/13)(1/(x-7)) dx#
#= -127/14 ln abs(x-8) -71/182ln abs(x+6)+97/13 ln abs(x-7) + C#
