Evaluate the following integrals:
`int frac{x^2}{(x^2+a^2)(x^2+b^2)}dx`

Answer

To evaluate the integral

\[ I = \int \frac{x^2}{(x^2 + a^2)(x^2 + b^2)} \, dx, \]

we will use the method of partial fractions.

### Step 1: Set up the partial fraction decomposition

We can express the integrand as:

\[ \frac{x^2}{(x^2 + a^2)(x^2 + b^2)} = \frac{A}{x^2 + a^2} + \frac{B}{x^2 + b^2}. \]

### Step 2: Multiply through by the denominator

Multiplying both sides by \((x^2 + a^2)(x^2 + b^2)\) gives:

\[ x^2 = A(x^2 + b^2) + B(x^2 + a^2). \]

### Step 3: Expand the right-hand side

Expanding the right-hand side, we have:

\[ x^2 = Ax^2 + Ab^2 + Bx^2 + Ba^2. \]

Combining like terms results in:

\[ x^2 = (A + B)x^2 + (Ab^2 + Ba^2). \]

### Step 4: Set up equations for coefficients

By comparing coefficients, we can set up the following equations:

1. \(A + B = 1\) (coefficient of \(x^2\))
2. \(Ab^2 + Ba^2 = 0\) (constant term)

### Step 5: Solve the system of equations

From the first equation, we can express \(B\) in terms of \(A\):

\[ B = 1 - A. \]

Substituting this into the second equation gives:

\[ A b^2 + (1 - A)a^2 = 0. \]

This simplifies to:

\[ A b^2 + a^2 - A a^2 = 0 \implies A(b^2 - a^2) = -a^2. \]

Thus,

\[ A = \frac{-a^2}{b^2 - a^2}. \]

Substituting \(A\) back into \(B = 1 - A\):

\[ B = 1 + \frac{a^2}{b^2 - a^2} = \frac{b^2 - a^2 + a^2}{b^2 - a^2} = \frac{b^2}{b^2 - a^2}. \]

### Step 6: Substitute back into the integral

Now we can rewrite the integral:

\[ I = \int \left( \frac{-a^2}{b^2 - a^2} \cdot \frac{1}{x^2 + a^2} + \frac{b^2}{b^2 - a^2} \cdot \frac{1}{x^2 + b^2} \right) dx. \]

### Step 7: Integrate each term

Now we can integrate each term separately:

\[ I = \frac{-a^2}{b^2 - a^2} \int \frac{1}{x^2 + a^2} \, dx + \frac{b^2}{b^2 - a^2} \int \frac{1}{x^2 + b^2} \, dx. \]

Using the formula \(\int \frac{1}{x^2 + c^2} \, dx = \frac{1}{c} \tan^{-1} \left( \frac{x}{c} \right) + C\), we find:

\[ I = \frac{-a^2}{b^2 - a^2} \cdot \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + \frac{b^2}{b^2 - a^2} \cdot \frac{1}{b} \tan^{-1} \left( \frac{x}{b} \right) + C. \]

### Step 8: Simplify the result

This simplifies to:

\[ I = \frac{-a}{b^2 - a^2} \tan^{-1} \left( \frac{x}{a} \right) + \frac{b}{b^2 - a^2} \tan^{-1} \left( \frac{x}{b} \right) + C. \]

### Final Result

Thus, the final result is:

\[ I = \frac{-a}{b^2 - a^2} \tan^{-1} \left( \frac{x}{a} \right) + \frac{b}{b^2 - a^2} \tan^{-1} \left( \frac{x}{b} \right) + C. \]