How do you integrate int (x-x^2)/((x+3)(x-1)(x+4)) using partial fractions?

1 Answer
George C.
Jun 11, 2017

int (x-x^2)/((x+3)(x-1)(x+4)) dx = 3 ln abs(x+3)-4 ln abs(x+4) + C

Explanation:

(x-x^2)/((x+3)(x-1)(x+4)) = A/(x+3)+B/(x-1)+C/(x+4)

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

A=((color(blue)(-3))-(color(blue)(-3))^2)/(((color(blue)(-3))-1)((color(blue)(-3))+4)) = (-12)/((-4)(1)) = 3

B=((color(blue)(1))-(color(blue)(1))^2)/(((color(blue)(1))+3)((color(blue)(1))+4)) = 0/((4)(5)) = 0

C=((color(blue)(-4))-(color(blue)(-4))^2)/(((color(blue)(-4))+3)((color(blue)(-4))-1)) = (-20)/((-1)(-5)) = -4

So:

int (x-x^2)/((x+3)(x-1)(x+4)) dx = int 3/(x+3)-4/(x+4) dx

color(white)(int (x-x^2)/((x+3)(x-1)(x+4)) dx) = 3 ln abs(x+3)-4 ln abs(x+4) + C

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