Perform the partial fraction decomposition of the following rational function: $\frac{3x^2-2x+1}{x^3-x^2}$.

Step 1 ：Step 1: To perform the partial fraction decomposition, we first factorize the denominator. Here, $x^3-x^2$ can be factorized to $x^2(x-1)$.

Step 2 ：Step 2: Once we have the factors, we can write the rational function as a sum of simpler fractions. We get: $\frac{3x^2-2x+1}{x^2(x-1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}$ where A, B, and C are constants to be determined.

Step 3 ：Step 3: Clearing the fractions, we get: $3x^2-2x+1=A(x^2)(x-1)+B(x)(x-1)+C(x^2)$.

Step 4 ：Step 4: Expanding and collecting like terms, we get: $3x^2-2x+1=(A+C)x^2+(-A+B)x+A$.

Step 5 ：Step 5: Comparing coefficients of the powers of x, we can set up three equations: $A+C=3$, $-A+B=-2$, and $A=1$.

Step 6 ：Step 6: Solving these equations, we find: $A=1$, $B=1$, and $C=2$.

Step 7 ：Step 7: Substituting the values of A, B, and C into the partial fractions, we get: $\frac{3x^2-2x+1}{x^3-x^2}=\frac{1}{x}+\frac{1}{x^2}+\frac{2}{x-1}$.