How do you integrate #int (3x^3 - 22x^2 + 27x + 46)/(x^2 - 8x + 15)# using partial fractions?

1 Answer
George C.
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:

#(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)#

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

#color(white)((3x^3-22x^2+27x+46)/(x^2-8x+15)) = 3x+2+(-2x+16)/((x-3)(x-5))#

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

We can determine #A# and #B# using Heaviside's cover up method...

#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#

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