We factorise the denominator
x^3-3x^2=x^2(x-3)
We start by performing a polynomial long division
color(white)(aaaa)x^3-3x^2color(white)(aaaaaa)-9color(white)(aaaa)|x^3-3x^2
color(white)(aaaa)x^3-3x^2color(white)(aaaaaa)#color(white)(aaaaaaa)|#1
color(white)(aaaaaa)0-0color(white)(aaaaaa)-9color(white)(aaaa)
Therefore,
(x^3-3x^2)/(x^3-3x^2)=1-9/(x^2(x-3))
Now, we perform the partial fraction decomposition
9/(x^2(x-3))=A/(x^2)+B/(x)+C/(x-3)
=(A(x-3)+B(x(x-3))+C(x^2))/(x^2(x-3))
The denominators are the same, we compare the numerators
9=A(x-3)+B(x(x-3))+C(x^2)
Let x=0,=>,9=-3A, =>, A=-3
Let x=3, =>, 9=9C, =>,C=1
Coefficients of x^2
0=B+C, =>, B=-C=-1
Therefore,
9/(x^2(x-3))=-3/(x^2)-1/(x)+1/(x-3)
So,
(x^3-3x^2)/(x^3-3x^2)=1-(-3/(x^2)-1/(x)+1/(x-3))
=1+3/(x^2)+1/(x)-1/(x-3)
Then,
int((x^3-3x^2)dx)/(x^3-3x^2)=int1dx+3intdx/(x^2)+intdx/(x)-intdx/(x-3)
=x-3/x+ln(|x|)-ln(|x-3|)+C
