The integral you've written can be interpreted as the following improper integral:

$\int_{-\infty}^{\infty} \frac{x^2}{9 + x^6} \, dx$

To solve this, observe the symmetry of the function, since the integrand is even (i.e., $f(x) = f(-x)$):

Symmetry Consideration:

The function $\frac{x^2}{9 + x^6}$ is even because both $x^2$ and $x^6$ contain only even powers of $x$. This allows us to simplify the integral as:

$2 \int_{0}^{\infty} \frac{x^2}{9 + x^6} \, dx$

Substitution:

Let's make a substitution to simplify the integral. Set:

$u = x^3 \quad \text{so that} \quad du = 3x^2 \, dx \quad \text{or} \quad dx = \frac{du}{3x^2} = \frac{du}{3u^{2/3}}$

Thus, the integral becomes:

$2 \int_{0}^{\infty} \frac{x^2}{9 + x^6} \, dx = 2 \int_{0}^{\infty} \frac{du}{3(9 + u^2)}$

This is now a standard arctangent integral. The result for such integrals is:

$\int \frac{du}{a^2 + u^2} = \frac{1}{a} \arctan\left(\frac{u}{a}\right)$

For our case, $a^2 = 9$ or $a = 3$, so the integral becomes:

$\frac{2}{9} \arctan\left(\frac{x^3}{3}\right) \Bigg|_0^\infty$

Evaluating the limits:

• As $x \to \infty$, $\arctan(\infty) = \frac{\pi}{2}$
• As $x = 0$, $\arctan(0) = 0$

So the final result is:

$\frac{2}{9} \times \frac{\pi}{2} = \frac{\pi}{9}$

Thus, the value of the integral is:

$\boxed{\frac{\pi}{9}}$

Would you like more details or have any questions?

Here are five related questions to explore:

1. How do improper integrals differ from regular integrals?
2. Why does symmetry help simplify definite integrals?
3. How does substitution aid in solving integrals involving higher powers?
4. What is the geometric interpretation of arctangent integrals?
5. How does one handle integrals with singularities or undefined points?

Tip: Recognizing symmetry in integrals can drastically reduce the complexity of the computation.