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.
