Question

How do you integrate `int (1-x^2)/((x-9)(x+3)(x-2))` using partial fractions?

Answer 1

`int (1-x^2)/((x-9)(x+3)(x-2)) dx`

`= -20/21 ln abs(x-9) -2/15 ln abs(x+3) + 3/35 ln abs(x-2) + C`

Explanation:

`(1-x^2)/((x-9)(x+3)(x-2)) = A/(x-9)+B/(x+3)+C/(x-2)`

Use Oliver Heaviside's cover up method to find:

`A = (1-(color(blue)(9))^2)/(((color(blue)(9))+3)((color(blue)(9))-2)) = (-80)/((12)(7)) = -20/21`

`B = (1-(color(blue)(-3))^2)/(((color(blue)(-3))-9)((color(blue)(-3))-2)) = (-8)/((-12)(-5)) = -2/15`

`C = (1-(color(blue)(2))^2)/(((color(blue)(2))-9)((color(blue)(2))+3)) = (-3)/((-7)(5)) = 3/35`

So:

`int (1-x^2)/((x-9)(x+3)(x-2)) dx`

`= int -20/21(1/(x-9))-2/15(1/(x+3))+3/35(1/(x-2)) dx`

`= -20/21 ln abs(x-9) -2/15 ln abs(x+3) + 3/35 ln abs(x-2) + C`