The denominator factors as: x(x^2-x+2) and cannot be factored further using real numbers (discriminant is negative). The integrand cannot be reduced, so we proceed:
Find, A, B, and C to make:
A/x +(Bx+C)/(x^2-x+2) = (x^2-5x+6)/(x(x^2-x+2))
We need:
Ax^2-Ax+2A+Bx^2+Cx = x^2-5x+6
Solve the system:
A+B=1
-A+C=-5
2A=6
Obviously, A=3 and this makes B=-2 and C=-2
The partial fraction decomposition is:
3/x -(2x+2)/(x^2-x+2) = (x^2-5x+6)/(x(x^2-x+2)).
If you have time check by combining the fractions on the left to make sure you've made no mistakes.
Now integrate:
int (x^2-5x+6)/(x^3-x^2+2x)) dx = int (3/x)dx - int (2x+2)/(x^2-x+2) dx
The first integral is easy. I'd split the second using:
(2x+2)/(x^2-x+2) = ((2x-1) +3)/(x^2-x+2) because
int (2x-1)/(x^2-x+2) dx is a simple substitution and then
int 3/(x^2-x+2) dx is an inverse tangent with some constants to figure out by completing the square and substituting.
