The partial fractions of `(x^(2))/((x^(2)+a^(2))(x^(2)+b^(2)))` are

Options

A. `(1)/(a^(2)+b^(2))[(a^(2))/(x^(2)+a^(2))-(b^(2))/(x^(2)+b^(2))]`
B. `(1)/(b^(2)-a^(2))[(a^(2))/(x^(2)+a^(2))-(b^(2))/(x^(2)+b^(2))]`
C. `(1)/(a^(2)-b^(2))[(a^(2))/(x^(2)+a^(2))-(b^(2))/(x^(2)+b^(2))]`
D. `(1)/(a^(2)-b^(2))[(1)/(x^(2)+a^(2))-(1)/(x^(2)+b^(2))]`

Answer

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

### Step 1: Set up the partial fraction decomposition
We start by expressing the given fraction in terms of its partial fractions:
\[ \frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = \frac{A}{x^2 + a^2} + \frac{B}{x^2 + b^2} \]
where \(A\) and \(B\) are constants that we need to determine.

### Step 2: Clear the denominators
Multiply both sides by \((x^2 + a^2)(x^2 + b^2)\) to eliminate the denominators:
\[ x^2 = A(x^2 + b^2) + B(x^2 + a^2) \]

### Step 3: Expand the right-hand side
Expanding the right-hand side gives:
\[ x^2 = Ax^2 + Ab^2 + Bx^2 + Ba^2 \]
Combining like terms, we have:
\[ x^2 = (A + B)x^2 + (Ab^2 + Ba^2) \]

### Step 4: Set up equations for coefficients
Now, we can equate the coefficients from both sides. The coefficient of \(x^2\) on the left side is 1, and the constant term is 0. Thus, we have the following system of equations:
1. \(A + B = 1\)
2. \(Ab^2 + Ba^2 = 0\)

### Step 5: Solve the system of equations
From the first equation, we can express \(B\) in terms of \(A\):
\[ B = 1 - A \]

Substituting \(B\) into the second equation:
\[ A b^2 + (1 - A) a^2 = 0 \]
This simplifies to:
\[ Ab^2 + a^2 - Aa^2 = 0 \]
Rearranging gives:
\[ A(b^2 - a^2) = -a^2 \]
Thus,
\[ A = \frac{-a^2}{b^2 - a^2} \]

Now substituting \(A\) back into the equation for \(B\):
\[ B = 1 - \frac{-a^2}{b^2 - a^2} = \frac{b^2 - a^2 + a^2}{b^2 - a^2} = \frac{b^2}{b^2 - a^2} \]

### Step 6: Write the partial fraction decomposition
Now we can substitute \(A\) and \(B\) back into the partial fraction form:
\[ \frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = \frac{-a^2}{b^2 - a^2} \cdot \frac{1}{x^2 + a^2} + \frac{b^2}{b^2 - a^2} \cdot \frac{1}{x^2 + b^2} \]

### Final Result
Thus, the partial fractions of \(\frac{x^2}{(x^2 + a^2)(x^2 + b^2)}\) are:
\[ \frac{-a^2}{b^2 - a^2} \cdot \frac{1}{x^2 + a^2} + \frac{b^2}{b^2 - a^2} \cdot \frac{1}{x^2 + b^2} \]