Since we're dividing by a single term, x^3,x3, we may simplify the integrand as follows:
(6x^5-2x^4+3x^3+x^2-x-2)/x^3=(6x^5)/x^3-(2x^4)/x^3+(3x^3)/x^3+x^2/x^3-x/x^3-2/x^36x5−2x4+3x3+x2−x−2x3=6x5x3−2x4x3+3x3x3+x2x3−xx3−2x3
=6x^2-2x+3+1/x-1/x^2-2/x^3=6x2−2x+3+1x−1x2−2x3
Thus, our integral becomes
int(6x^2-2x+3+1/x-1/x^2-2/x^3)dx∫(6x2−2x+3+1x−1x2−2x3)dx
=int(6x^2-2x+3+1/x-x^-2-2x^-3)dx=∫(6x2−2x+3+1x−x−2−2x−3)dx
Integrate, recalling that
intx^adx=x^(a+1)/(a+1)+C, a ne -1, int1/xdx=ln|x|+C∫xadx=xa+1a+1+C,a≠−1,∫1xdx=ln|x|+C
=int(6x^2-2x+3+1/x-x^-2-2x^-3)dx=6/3x^3-2/2x^2+3x+ln|x|+x^-1+2/2x^-2+C=∫(6x2−2x+3+1x−x−2−2x−3)dx=63x3−22x2+3x+ln|x|+x−1+22x−2+C
=2x^3-x^2+3x+ln|x|+1/x+1/x^2+C=2x3−x2+3x+ln|x|+1x+1x2+C
