How do you integrate int (3x^3 - 22x^2 + 27x + 46)/(x^2 - 8x + 15)3x322x2+27x+46x28x+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) + C3x322x2+27x+46x28x+15dx=32x2+2x5ln|x3|+3ln|x5|+C

Explanation:

(3x^3-22x^2+27x+46)/(x^2-8x+15) = ((3x^3-24x^2+45x)+(2x^2-16x+30)+(-2x+16))/(x^2-8x+15)3x322x2+27x+46x28x+15=(3x324x2+45x)+(2x216x+30)+(2x+16)x28x+15

color(white)((3x^3-22x^2+27x+46)/(x^2-8x+15)) = 3x+2+(-2x+16)/(x^2-8x+15)3x322x2+27x+46x28x+15=3x+2+2x+16x28x+15

color(white)((3x^3-22x^2+27x+46)/(x^2-8x+15)) = 3x+2+(-2x+16)/((x-3)(x-5))3x322x2+27x+46x28x+15=3x+2+2x+16(x3)(x5)

color(white)((3x^3-22x^2+27x+46)/(x^2-8x+15)) = 3x+2+A/(x-3)+B/(x-5)3x322x2+27x+46x28x+15=3x+2+Ax3+Bx5

We can determine AA and BB using Heaviside's cover up method...

A = (-2(color(blue)(3))+16)/(color(blue)(3)-5) = 10/(-2) = -5A=2(3)+1635=102=5

B = (-2(color(blue)(5))+16)/(color(blue)(5)-3) = 6/2 = 3B=2(5)+1653=62=3

So:

int (3x^3-22x^2+27x+46)/(x^2-8x+15) dx = int 3x+2-5/(x-3)+3/(x-5) dx3x322x2+27x+46x28x+15dx=3x+25x3+3x5dx

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) + C3x322x2+27x+46x28x+15dx=32x2+2x5ln|x3|+3ln|x5|+C