The integral provided is:

$\int_1^{\infty} \frac{x^2}{(x^3 + 2)^2} \, dx$

Let's solve this step by step.

Step 1: Substitution

We will use a substitution to simplify the integrand. Let:

$u = x^3 + 2$

Then, the derivative of $u$ with respect to $x$ is:

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

Thus, $x^2 \, dx = \frac{du}{3}$.

Now, change the limits of integration. When $x = 1$, $u = 1^3 + 2 = 3$, and when $x \to \infty$, $u \to \infty$.

So the integral becomes:

$\int_3^{\infty} \frac{1}{3u^2} \, du$

Step 2: Simplify and Integrate

Now we can integrate this expression. The integral of $u^{-2}$ is:

$\int u^{-2} \, du = -u^{-1}$

Therefore, the integral becomes:

$\frac{1}{3} \left[ -\frac{1}{u} \right]_3^{\infty}$

Step 3: Evaluate the Limits

Evaluating this at the limits:

$\frac{1}{3} \left( -\frac{1}{\infty} + \frac{1}{3} \right) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$

Final Answer:

$\int_1^{\infty} \frac{x^2}{(x^3 + 2)^2} \, dx = \frac{1}{9}$

Would you like more details or have any questions? Here are some related problems to practice:

1. Evaluate $\int_1^{\infty} \frac{x^4}{(x^5 + 1)^2} \, dx$.
2. Solve $\int_0^{\infty} \frac{dx}{(x^2 + 1)^2}$.
3. Compute $\int_1^{\infty} \frac{x}{(x^2 + 1)^2} \, dx$.
4. Find the value of $\int_2^{\infty} \frac{dx}{(x^3 + 4)^2}$.
5. Evaluate $\int_0^{1} \frac{x^2}{(x^3 + 2)^2} \, dx$.

Tip: When solving improper integrals, always check if the integrand converges at infinity to ensure the integral has a finite value.