The partial fractions of `(9)/((x-1)(x+2)^2)` are ………

Answer

To find the partial fractions of the expression \(\frac{9}{(x-1)(x+2)^2}\), we will follow these steps:

### Step 1: Set up the partial fraction decomposition
We start by expressing the given fraction in terms of its partial fractions. Since the denominator has a linear factor \((x-1)\) and a repeated linear factor \((x+2)^2\), we can write:

\[ \frac{9}{(x-1)(x+2)^2} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2} \]

where \(A\), \(B\), and \(C\) are constants that we need to determine.

### Step 2: Clear the denominators
To eliminate the denominators, we multiply both sides by the common denominator \((x-1)(x+2)^2\):

\[ 9 = A(x+2)^2 + B(x-1)(x+2) + C(x-1) \]

### Step 3: Expand the right-hand side
Now, we will expand the right-hand side:

1. **Expand \(A(x+2)^2\)**:
\[ A(x+2)^2 = A(x^2 + 4x + 4) = Ax^2 + 4Ax + 4A \]

2. **Expand \(B(x-1)(x+2)\)**:
\[ B(x-1)(x+2) = B(x^2 + x - 2) = Bx^2 + Bx - 2B \]

3. **Expand \(C(x-1)\)**:
\[ C(x-1) = Cx - C \]

Combining these expansions, we have:

\[ 9 = (A + B)x^2 + (4A + B + C)x + (4A - 2B - C) \]

### Step 4: Set up equations by comparing coefficients
Now we can compare coefficients from both sides of the equation:

1. For \(x^2\): \(A + B = 0\) (Equation 1)
2. For \(x\): \(4A + B + C = 0\) (Equation 2)
3. For the constant term: \(4A - 2B - C = 9\) (Equation 3)

### Step 5: Solve the system of equations
From Equation 1, we have:

\[ B = -A \]

Substituting \(B = -A\) into Equations 2 and 3:

**Substituting into Equation 2**:
\[ 4A - A + C = 0 \implies 3A + C = 0 \implies C = -3A \quad (Equation 4) \]

**Substituting into Equation 3**:
\[ 4A - 2(-A) - (-3A) = 9 \implies 4A + 2A + 3A = 9 \implies 9A = 9 \implies A = 1 \]

Now substituting \(A = 1\) back into Equations 1 and 4:

1. From Equation 1:
\[ B = -1 \]
2. From Equation 4:
\[ C = -3 \]

### Step 6: Write the final partial fraction decomposition
Now that we have \(A\), \(B\), and \(C\), we can write the partial fractions:

\[ \frac{9}{(x-1)(x+2)^2} = \frac{1}{x-1} - \frac{1}{x+2} - \frac{3}{(x+2)^2} \]

### Final Answer
The partial fractions of \(\frac{9}{(x-1)(x+2)^2}\) are:

\[ \frac{1}{x-1} - \frac{1}{x+2} - \frac{3}{(x+2)^2} \]

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