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The given improper integral is:

\[
\int_{1}^{\infty} \frac{x^2 \sin^2(x)}{x^4 + 2x^2 + 7} \, dx
\]

To determine whether this integral converges or diverges, we will use the **Comparison Test for Improper Integrals**. The key idea is to compare this integral to a simpler function that is easier to analyze and behaves similarly to the given function for large values of \( x \).

### Step 1: Analyze the behavior of the function for large \( x \)

For large \( x \), the denominator \( x^4 + 2x^2 + 7 \) behaves like \( x^4 \), because the \( x^4 \) term dominates the other terms. Similarly, the numerator contains \( x^2 \sin^2(x) \), where \( \sin^2(x) \) oscillates between 0 and 1. Hence, the given function is approximately:

\[
\frac{x^2 \sin^2(x)}{x^4} = \frac{\sin^2(x)}{x^2} \text{ for large } x.
\]

This suggests that the function we are working with behaves similarly to \( \frac{\sin^2(x)}{x^2} \) as \( x \to \infty \).

### Step 2: Set up a comparison function

The function \( \frac{\sin^2(x)}{x^2} \) oscillates between 0 and \( \frac{1}{x^2} \), so we can compare the given function to \( \frac{1}{x^2} \), which is simpler to analyze.

Consider the integral:

\[
\int_{1}^{\infty} \frac{1}{x^2} \, dx.
\]

This is a standard p-integral with \( p = 2 \), and we know that:

\[
\int_{1}^{\infty} \frac{1}{x^2} \, dx = 1 \text{ (which converges)}.
\]

### Step 3: Apply the Comparison Test

Since \( 0 \leq \frac{\sin^2(x)}{x^2} \leq \frac{1}{x^2} \) for all \( x \geq 1 \), and the integral of \( \frac{1}{x^2} \) converges, by the Comparison Test, the given integral:

\[
\int_{1}^{\infty} \frac{x^2 \sin^2(x)}{x^4 + 2x^2 + 7} \, dx
\]

also **converges**.

### Conclusion:

The improper integral converges based on the Comparison Test.

---

Would you like more details on any part of this solution? Here are 5 questions that expand on this concept:

1. Why does the \( x^4 \) term dominate the behavior of the denominator for large \( x \)?
2. How does the oscillation of \( \sin^2(x) \) affect the integral's behavior at infinity?
3. What are the conditions under which the Comparison Test applies to improper integrals?
4. How would the result change if the numerator had a different power of \( x \)?
5. Can the Limit Comparison Test be used as an alternative to the direct Comparison Test here?

**Tip**: When analyzing improper integrals, focusing on the behavior of the integrand for large values of \( x \) is key to choosing a comparison function.

Use the Comparison Test for Improper Integrals to determine whether the integral converges or diverges: ∫[1, ∞] (x² sin²(x)) / (x⁴ + 2x² + 7) dx

This detailed solution explains how to determine whether the improper integral ∫[1, ∞] (x² sin²(x)) / (x⁴ + 2x² + 7) dx converges using the Comparison Test. Key concepts include improper integrals, asymptotic behavior, and the comparison test. The problem is suited for undergraduate calculus students.

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