Answer
$\frac{3}{x}+\frac{4}{x^2}-\frac{3}{x-1}$
Work Step by Step
1. Let $\frac{x-4}{x^2(x-1)}=\frac{A}{x-1}+\frac{B}{x}+\frac{C}{x^2}$
2. Combine the right side to get $\frac{Ax^2+Bx^2-Bx+Cx-C}{x^2(x-1)}=\frac{(A+B)x^2-(B-C)x-C}{x^2(x-1)}$
3. Compare with step 1, we have $A+B=0, B-C=-1, C=4$, thus $B=3$ and $A=-3$
4. We have $\frac{x-4}{x^2(x-1)}=\frac{3}{x}+\frac{4}{x^2}-\frac{3}{x-1}$
