How do you integrate #int (3x^3 - 22x^2 + 27x + 46)/(x^2 - 8x + 15)# using partial fractions?
1 Answer
Oct 2, 2016
#int (3x^3-22x^2+27x+46)/(x^2-8x+15) dx = 3/2x^2+2x-5ln abs(x-3)+3ln abs(x-5) + C#
Explanation:
We can determine
#A = (-2(color(blue)(3))+16)/(color(blue)(3)-5) = 10/(-2) = -5#
#B = (-2(color(blue)(5))+16)/(color(blue)(5)-3) = 6/2 = 3#
So:
#int (3x^3-22x^2+27x+46)/(x^2-8x+15) dx = int 3x+2-5/(x-3)+3/(x-5) dx#
#color(white)(int (3x^3-22x^2+27x+46)/(x^2-8x+15) dx) = 3/2x^2+2x-5ln abs(x-3)+3ln abs(x-5) + C#