How do you find #int x/(x^2-x-2)dx# using partial fractions?
1 Answer
Oct 2, 2016
Explanation:
#x/(x^2-x-2) = x/((x-2)(x+1)) = 2/(3(x-2))+1/(3(x+1))#
So:
#int x/(x^2-x-2) dx = int 2/(3(x-2))+1/(3(x+1)) dx#
#color(white)(int x/(x^2-x-2) dx) = 2/3 ln abs (x-2)+1/3 ln abs(x+1) + C#
