How do you integrate int (3x^3 - 22x^2 + 27x + 46)/(x^2 - 8x + 15)∫3x3−22x2+27x+46x2−8x+15 using partial fractions?
1 Answer
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∫3x3−22x2+27x+46x2−8x+15dx=32x2+2x−5ln|x−3|+3ln|x−5|+C
Explanation:
We can determine
A = (-2(color(blue)(3))+16)/(color(blue)(3)-5) = 10/(-2) = -5A=−2(3)+163−5=10−2=−5
B = (-2(color(blue)(5))+16)/(color(blue)(5)-3) = 6/2 = 3B=−2(5)+165−3=62=3
So:
int (3x^3-22x^2+27x+46)/(x^2-8x+15) dx = int 3x+2-5/(x-3)+3/(x-5) dx∫3x3−22x2+27x+46x2−8x+15dx=∫3x+2−5x−3+3x−5dx
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∫3x3−22x2+27x+46x2−8x+15dx=32x2+2x−5ln|x−3|+3ln|x−5|+C