Question

How do you integrate `int (1-x^2)/((x-9)(x+6)(x-4))` using partial fractions?

Answer 1

`int (1-x^2)/((x-9)(x+6)(x-4)) dx =`

`=-16/15 ln abs(x-9) - 7/30 ln abs(x+6) + 3/10 ln abs(x-4) + C`

Explanation:

`(1-x^2)/((x-9)(x+6)(x-4)) = A/(x-9)+B/(x+6)+C/(x-4)`

We can determine `A`, `B` and `C` using Heaviside's cover up method:

`A = (1-color(blue)(9)^2)/((color(blue)(9)+6)(color(blue)(9)-4)) = (-80)/((15)(5)) = -16/15`

`B = (1-color(blue)((-6))^2)/((color(blue)((-6))-9)(color(blue)((-6))-4)) = (-35)/((-15)(-10)) = -7/30`

`C = (1-color(blue)(4)^2)/((color(blue)(4)-9)(color(blue)(4)+6)) = (-15)/((-5)(10)) = 3/10`

So:

`int (1-x^2)/((x-9)(x+6)(x-4)) dx =`

`int -16/15(1/(x-9)) - 7/30(1/(x+6))+3/10(1/(x-4)) dx`

`=-16/15 ln abs(x-9) - 7/30 ln abs(x+6) + 3/10 ln abs(x-4) + C`

where `C` is the constant of integeration (not to be confused with the coefficient `C` we calculated earlier).