To decompose the rational expression $\frac{x}{(x-1)^2(x-2)}$ into partial fractions, we'll express it in the following form:

$\frac{x}{(x-1)^2(x-2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x-2}$

Here, $A$, $B$, and $C$ are constants that we need to determine.

Step 1: Set up the equation

Multiply both sides by the denominator $(x-1)^2(x-2)$ to eliminate the fractions:

$x = A(x-1)(x-2) + B(x-2) + C(x-1)^2$

Step 2: Expand the right-hand side

Let's expand each term on the right-hand side:

$A(x-1)(x-2) = A(x^2 - 3x + 2)$ $B(x-2) = Bx - 2B$ $C(x-1)^2 = C(x^2 - 2x + 1)$

Thus, the equation becomes:

$x = A(x^2 - 3x + 2) + B(x - 2) + C(x^2 - 2x + 1)$

Step 3: Combine like terms

Combine the like terms to group them by powers of $x$:

$x = (A + C)x^2 + (-3A + B - 2C)x + (2A - 2B + C)$

Step 4: Compare coefficients

Now, compare the coefficients of $x^2$, $x$, and the constant term on both sides of the equation:

1. Coefficient of $x^2$: $0 = A + C$
2. Coefficient of $x$: $1 = -3A + B - 2C$
3. Constant term: $0 = 2A - 2B + C$

Step 5: Solve the system of equations

We now have the following system of linear equations:

$A + C = 0 \quad \text{(Equation 1)}$ $-3A + B - 2C = 1 \quad \text{(Equation 2)}$ $2A - 2B + C = 0 \quad \text{(Equation 3)}$

From Equation 1, we know $C = -A$.

Substitute $C = -A$ into Equations 2 and 3:

$-3A + B - 2(-A) = 1 \quad \Rightarrow \quad -3A + B + 2A = 1 \quad \Rightarrow \quad -A + B = 1 \quad \text{(Equation 4)}$

$2A - 2B - A = 0 \quad \Rightarrow \quad A - 2B = 0 \quad \Rightarrow \quad A = 2B \quad \text{(Equation 5)}$

Step 6: Find $A$, $B$, and $C$

Using Equation 4 and 5:

Substitute $A = 2B$ into Equation 4:

$-2B + B = 1 \quad \Rightarrow \quad -B = 1 \quad \Rightarrow \quad B = -1$

Thus:

$A = 2B = 2(-1) = -2$

And:

$C = -A = 2$

Final Answer:

The partial fraction decomposition is:

$\frac{x}{(x-1)^2(x-2)} = \frac{-2}{x-1} + \frac{-1}{(x-1)^2} + \frac{2}{x-2}$

Would you like a detailed explanation of any part of the process? Here are some related questions:

1. What is the purpose of partial fraction decomposition?
2. Can partial fractions be used with non-linear terms?
3. How do you handle complex roots in partial fraction decomposition?
4. How do partial fractions apply to solving differential equations?
5. How do you perform partial fraction decomposition when the numerator's degree is greater than the denominator's?
6. Can you perform partial fractions on trigonometric functions?
7. How do you verify the correctness of a partial fraction decomposition?
8. What are some common applications of partial fractions in calculus?

Tip: Always check your partial fraction decomposition by combining the fractions and simplifying to see if you return to the original expression.