find the integral using Integration by partial Fractions Here is my work for this problem...just wanted a check over and see if i missed anything
Original Problem: $\int$ $\frac6{x^3-3x^2}$
F 6/x^3-3x^2= F 6/x^2(x-3)
6/x^2(x-3)= Ax+B/x^2+C/(x-3)
C/x^2(x-3)= (A+C)x^2+(B-3A)x-3B
B=-2
A=-2/3
C=1/3
F C/x^2-3x^2 dx= F(-2/3)x-2/x^2+ F 2/3(x-3)
F -2/3x-2/x^2+2/3(x-3)
F 6/x^3-3x^2=-2/3 ln (x)+2/x+2/3 ln(x-3)

We have the expansion
$$\frac{1}{x^3-3x^2}=\frac{A+Bx}{x^2}+\frac{C}{x-3}\tag 1$$
Multiplying both sides of $(1)$ by $x-3$ and letting $x\to 3$ reveals that $C=1/9$.  
Multiplying both sides of $(1)$ by $x^2$ and letting $x\to 0$ reveals that $A=-1/3$.
Multiplying both sides of $(1)$ by $x^2$, taking a derivative with respect to $x$, and letting $x\to 0$ reveals that $B=-1/9$.
Thus, we have 
$$\begin{align}\int\frac{6}{x^3-3x^2}\,dx&=\int\left(\frac{-2}{x^2}+\frac{-2/3}{x}+\frac{2/3}{x-3}\right)dx\\\\
&=\frac{2}{x}-\frac23\log |x|+\frac23 \log |x-3|+C
\end{align}$$
