To decompose the expression $\frac{x}{(x-1)^2(x-2)}$ into partial fractions, let's follow the steps below.

Step 1: Set Up the Partial Fraction Decomposition

Since the denominator $(x-1)^2(x-2)$ contains a repeated linear factor $(x-1)^2$ and a distinct linear factor $(x-2)$, the partial fraction decomposition will take the form:

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

Step 2: Clear the Denominator

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

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

Step 3: Expand and Simplify

Expand the right side of the equation:

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

Distribute the terms:

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

Combine like terms:

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

Step 4: Equate Coefficients

For the equation to hold true for all $x$, the coefficients of like terms on both sides must be equal. This gives us a system of equations:

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

Step 5: Solve the System of Equations

Let's solve the system:

1. From the first equation, $A + C = 0$ $\Rightarrow C = -A$.

2. Substitute $C = -A$ into the second equation:

   $-3A - 2(-A) + B = 1 \implies -3A + 2A + B = 1 \implies -A + B = 1 \implies B = A + 1$

3. Substitute $C = -A$ and $B = A + 1$ into the third equation:

   $2A - 2(A + 1) - A = 0 \implies 2A - 2A - 2 - A = 0 \implies -A - 2 = 0 \implies A = -2$

4. Now, find $B$ and $C$:

   $B = A + 1 = -2 + 1 = -1$

   $C = -A = 2$

Step 6: Write the Final Partial Fraction Decomposition

Substituting $A = -2$, $B = -1$, and $C = 2$ back into the partial fractions:

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

This simplifies to:

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

Would you like further details on any of these steps or have any questions?

Related Questions

1. How can you solve partial fraction decompositions with higher powers in the denominator?
2. What is the difference between partial fraction decomposition for distinct linear factors and repeated linear factors?
3. How do you integrate a function after performing partial fraction decomposition?
4. Can partial fractions be used when the denominator has irreducible quadratic factors?
5. How do you perform partial fraction decomposition for a rational function with complex roots?
6. What is the method to verify the correctness of partial fraction decomposition?
7. How does the degree of the numerator affect the partial fraction decomposition?
8. What are some common applications of partial fraction decomposition in differential equations?

Tip

When performing partial fraction decomposition, always start by equating the coefficients of corresponding powers of $x$ on both sides of the equation to form a system of linear equations. Solving this system will give you the values of the constants in the partial fraction.