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

Step 1: Simplify the Expression

The given expression is:

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

This simplifies to:

$\frac{x-1}{(x-2)^3}$

Step 2: Set Up the Partial Fraction Decomposition

We want to express $\frac{x-1}{(x-2)^3}$ as a sum of simpler fractions. Since the denominator has a cubic factor $(x-2)^3$, the partial fraction decomposition will take the form:

$\frac{x-1}{(x-2)^3} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)^3}$

Step 3: Clear the Denominator

Multiply both sides by $(x-2)^3$ to clear the denominator:

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

Step 4: Expand and Simplify

Expand the right side of the equation:

$x - 1 = A(x^2 - 4x + 4) + B(x - 2) + C$

Distribute the terms:

$x - 1 = A(x^2 - 4x + 4) + Bx - 2B + C$

Combine like terms:

$x - 1 = Ax^2 - 4Ax + 4A + Bx - 2B + C$

Step 5: 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 $x^2$: $A = 0$
2. Coefficient of $x$: $-4A + B = 1$
3. Constant term: $4A - 2B + C = -1$

Step 6: Solve the System of Equations

From the first equation, $A = 0$.

Substitute $A = 0$ into the second equation:

$-4(0) + B = 1 \implies B = 1$

Substitute $A = 0$ and $B = 1$ into the third equation:

$4(0) - 2(1) + C = -1 \implies -2 + C = -1 \implies C = 1$

Step 7: Write the Final Partial Fraction Decomposition

Substituting $A = 0$, $B = 1$, and $C = 1$ back into the partial fractions:

$\frac{x-1}{(x-2)^3} = \frac{1}{(x-2)^2} + \frac{1}{(x-2)^3}$

So, the partial fraction decomposition of $\frac{x-1}{(x-2)^3}$ is:

$\frac{x-1}{(x-2)^3} = \frac{1}{(x-2)^2} + \frac{1}{(x-2)^3}$

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

Related Questions

1. How do you perform partial fraction decomposition for a quadratic denominator?
2. What are the conditions for using partial fractions in calculus?
3. How do you decompose a rational function with repeated linear factors?
4. Can partial fractions be used for improper fractions? How?
5. What is the method to solve partial fractions when the denominator has complex roots?
6. How do you handle partial fraction decomposition when the numerator is not of lower degree than the denominator?
7. How can partial fractions be applied in integration problems?
8. What are the common errors to avoid in partial fraction decomposition?

Tip

Always check if the fraction is proper (the degree of the numerator is less than the degree of the denominator) before attempting partial fraction decomposition. If not, perform polynomial division first.