Evaluate: `int(x^2)/((x^2+a^2)(x^2+b^2))dx`

Explanation

To evaluate the integral

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

we can use the method of partial fractions. Here’s a step-by-step solution:

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

We want to express

\[ \frac{x^2}{(x^2 + a^2)(x^2 + b^2)} \]

as a sum of simpler fractions. We can write:

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

for some constants \(A\) and \(B\).

### Step 2: Clear the denominators

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 and collect like terms

Expanding the right-hand side:

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

Combining the \(x^2\) terms:

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

### Step 4: Set up equations by comparing coefficients

Now, we can compare coefficients from both sides:

1. For \(x^2\): \(A + B = 1\)
2. For the constant term: \(Ab^2 + Ba^2 = 0\)

### Step 5: Solve the system of equations

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

\[ Ab^2 + Ba^2 = 0 \implies A = -\frac{Ba^2}{b^2}. \]

Substituting this into the first equation:

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

Now substituting \(B\) back to find \(A\):

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

### Step 6: Rewrite the integral

Now substituting \(A\) and \(B\) back into 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:

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

Using the formula

\[ \int \frac{1}{x^2 + c^2} \, dx = \frac{1}{c} \tan^{-1}\left(\frac{x}{c}\right) + C, \]

we have:

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

### Step 8: Simplify the expression

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 answer is:

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

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