How do you integrate #(x^2+x+1)/(1-x^2)# using partial fractions?
1 Answer
Oct 10, 2016
Explanation:
#(x^2+x+1)/(1-x^2) = (-x^2-x-1)/(x^2-1)#
#color(white)((x^2+x+1)/(1-x^2)) = (-(x^2-1)-x-2)/(x^2-1)#
#color(white)((x^2+x+1)/(1-x^2)) = -1+(-x-2)/(x^2-1)#
#color(white)((x^2+x+1)/(1-x^2)) = -1+(-x-2)/((x-1)(x+1))#
#color(white)((x^2+x+1)/(1-x^2)) = -1+A/(x-1)+B/(x+1)#
Use Heaviside's cover up method to find:
#A = (-(color(blue)(1))-2)/((color(blue)(1))+1) = -3/2#
#B = (-(color(blue)(-1))-2)/((color(blue)(-1))-1) = (-1)/(-2) = 1/2#
So:
#int (x^2+x+1)/(1-x^2) dx = int -1-3/2(1/(x-1))+1/2(1/(x+1)) dx#
#color(white)(int (x^2+x+1)/(1-x^2) dx) = -x-3/2ln abs(x-1)+1/2 ln abs (x+1) + C#
