The partial fractions of `(1)/((x^(2)+9)(x^(2)+16))` are

Options

A. `1/7 [ (1)/(x^(2)+9)-(1)/(x^(2)+16)]`
B. `1/9 [ (1)/(x^(2)+9)-(1)/(x^(2)+16)]`
C. `1/7 [ (1)/(x^(2)+16)-(1)/(x^(2)+9)]`
D. `1/25 [ (1)/(x^(2)+9)-(1)/(x^(2)+16)]`

Answer

To find the partial fractions of the expression \(\frac{1}{(x^2 + 9)(x^2 + 16)}\), we can follow these steps:

### Step 1: Set Up the Partial Fraction Decomposition
We start by expressing the given fraction as a sum of simpler fractions:
\[ \frac{1}{(x^2 + 9)(x^2 + 16)} = \frac{A}{x^2 + 9} + \frac{B}{x^2 + 16} \]
where \(A\) and \(B\) are constants that we need to determine.

### Step 2: Clear the Denominator
To eliminate the denominators, we multiply both sides by \((x^2 + 9)(x^2 + 16)\):
\[ 1 = A(x^2 + 16) + B(x^2 + 9) \]

### Step 3: Expand the Right Side
Now we expand the right-hand side:
\[ 1 = Ax^2 + 16A + Bx^2 + 9B \]
This simplifies to:
\[ 1 = (A + B)x^2 + (16A + 9B) \]

### Step 4: Set Up the System of Equations
Now we can equate the coefficients from both sides. Since the left side has no \(x^2\) term, we have:
1. \(A + B = 0\) (coefficient of \(x^2\))
2. \(16A + 9B = 1\) (constant term)

### Step 5: Solve the System of Equations
From the first equation, we can express \(B\) in terms of \(A\):
\[ B = -A \]
Substituting \(B = -A\) into the second equation:
\[ 16A + 9(-A) = 1 \implies 16A - 9A = 1 \implies 7A = 1 \implies A = \frac{1}{7} \]
Now substituting \(A\) back to find \(B\):
\[ B = -\frac{1}{7} \]

### Step 6: Write the Partial Fraction Decomposition
Now that we have \(A\) and \(B\), we can write the partial fraction decomposition:
\[ \frac{1}{(x^2 + 9)(x^2 + 16)} = \frac{1/7}{x^2 + 9} - \frac{1/7}{x^2 + 16} \]
Factoring out \(\frac{1}{7}\):
\[ = \frac{1}{7} \left( \frac{1}{x^2 + 9} - \frac{1}{x^2 + 16} \right) \]

### Final Answer
Thus, the partial fractions of \(\frac{1}{(x^2 + 9)(x^2 + 16)}\) are:
\[ \frac{1}{7} \left( \frac{1}{x^2 + 9} - \frac{1}{x^2 + 16} \right) \]

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