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.