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

1 Answer
George C.
Sep 10, 2016

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).

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