How do you integrate int (x-x^2)/((x+3)(x-1)(x+4)) ∫x−x2(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∫x−x2(x+3)(x−1)(x+4)dx=3ln|x+3|−4ln|x+4|+C

Explanation:

(x-x^2)/((x+3)(x-1)(x+4)) = A/(x+3)+B/(x-1)+C/(x+4)x−x2(x+3)(x−1)(x+4)=Ax+3+Bx−1+Cx+4

We can find AA, BB and CC 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)) = 3A=(−3)−(−3)2((−3)−1)((−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)) = 0B=(1)−(1)2((1)+3)((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)) = -4C=(−4)−(−4)2((−4)+3)((−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∫x−x2(x+3)(x−1)(x+4)dx=∫3x+3−4x+4dx

color(white)(int (x-x^2)/((x+3)(x-1)(x+4)) dx) = 3 ln abs(x+3)-4 ln abs(x+4) + C∫x−x2(x+3)(x−1)(x+4)dx=3ln|x+3|−4ln|x+4|+C

Related questions
See all questions in Integral by Partial Fractions
Impact of this question
1260 views around the world
You can reuse this answer
Creative Commons License