Find is the coefficient of `x^(4)` in the power series expansion of `(3x^(2)+2x)/((x^(2)+2)(x-3))` specifying the interval in which the expansion is valid.

Answer

To find the coefficient of \( x^4 \) in the power series expansion of \( \frac{3x^2 + 2x}{(x^2 + 2)(x - 3)} \), we will follow these steps:

### Step 1: Partial Fraction Decomposition
We start by expressing the given function in terms of partial fractions:
\[ \frac{3x^2 + 2x}{(x^2 + 2)(x - 3)} = \frac{Ax + B}{x^2 + 2} + \frac{C}{x - 3} \]
where \( A \), \( B \), and \( C \) are constants to be determined.

### Step 2: Combine the Right Side
We will combine the right-hand side over a common denominator:
\[ \frac{Ax + B}{x^2 + 2} + \frac{C}{x - 3} = \frac{(Ax + B)(x - 3) + C(x^2 + 2)}{(x^2 + 2)(x - 3)} \]

### Step 3: Equate Numerators
Now, we equate the numerators:
\[ 3x^2 + 2x = (Ax + B)(x - 3) + C(x^2 + 2) \]
Expanding the right side:
\[ (Ax + B)(x - 3) = Ax^2 - 3Ax + Bx - 3B \]
Thus, we have:
\[ 3x^2 + 2x = Ax^2 - 3Ax + Bx - 3B + Cx^2 + 2C \]
Combining like terms gives:
\[ 3x^2 + 2x = (A + C)x^2 + (-3A + B)x + (-3B + 2C) \]

### Step 4: Set Up the System of Equations
From the equality of coefficients, we have the following equations:
1. \( A + C = 3 \)
2. \( -3A + B = 2 \)
3. \( -3B + 2C = 0 \)

### Step 5: Solve the System of Equations
From equation (1), we can express \( C \) in terms of \( A \):
\[ C = 3 - A \]
Substituting \( C \) into equation (3):
\[ -3B + 2(3 - A) = 0 \implies -3B + 6 - 2A = 0 \implies 3B = 6 - 2A \implies B = \frac{6 - 2A}{3} \]
Now substituting \( B \) into equation (2):
\[ -3A + \frac{6 - 2A}{3} = 2 \]
Multiplying through by 3 to eliminate the fraction:
\[ -9A + 6 - 2A = 6 \implies -11A = 0 \implies A = 0 \]
Substituting \( A = 0 \) back into the equations:
\[ C = 3 - 0 = 3 \quad \text{and} \quad B = \frac{6 - 2(0)}{3} = 2 \]

### Step 6: Write the Partial Fraction Decomposition
Thus, we have:
\[ \frac{3x^2 + 2x}{(x^2 + 2)(x - 3)} = \frac{2}{x^2 + 2} + \frac{3}{x - 3} \]

### Step 7: Power Series Expansion
Next, we will find the power series expansion for each term:
1. For \( \frac{2}{x^2 + 2} \):
\[ \frac{2}{x^2 + 2} = \frac{2}{2(1 + \frac{x^2}{2})} = \frac{1}{1 + \frac{x^2}{2}} = 1 - \frac{x^2}{2} + \frac{x^4}{4} - \cdots \]
2. For \( \frac{3}{x - 3} \):
\[ \frac{3}{x - 3} = -\frac{3}{3(1 - \frac{x}{3})} = -1 \cdot \frac{1}{1 - \frac{x}{3}} = -1 \left(1 + \frac{x}{3} + \frac{x^2}{9} + \frac{x^3}{27} + \frac{x^4}{81} + \cdots\right) \]

### Step 8: Combine the Series
Now we combine the series:
\[ \frac{3x^2 + 2x}{(x^2 + 2)(x - 3)} = \left(1 - \frac{x^2}{2} + \frac{x^4}{4} - \cdots\right) - \left(1 + \frac{x}{3} + \frac{x^2}{9} + \frac{x^3}{27} + \frac{x^4}{81} + \cdots\right) \]

### Step 9: Collect Coefficients of \( x^4 \)
The coefficient of \( x^4 \) from \( \frac{2}{x^2 + 2} \) is \( \frac{1}{4} \) and from \( \frac{3}{x - 3} \) is \( -\frac{1}{81} \):
\[ \text{Coefficient of } x^4 = \frac{1}{4} - \frac{1}{81} \]
Finding a common denominator (which is 324):
\[ \frac{1}{4} = \frac{81}{324}, \quad -\frac{1}{81} = -\frac{4}{324} \]
Thus,
\[ \frac{81 - 4}{324} = \frac{77}{324} \]

### Final Answer
The coefficient of \( x^4 \) in the power series expansion is \( \frac{77}{324} \).