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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.

Evaluate the integral \(\int_1^{\infty} \frac{x^2}{(x^3 + 2)^2} \, dx\).

Explore how to evaluate the improper integral \(\int_1^{\infty} \frac{x^2}{(x^3 + 2)^2} \, dx\) using substitution techniques and understanding integral convergence. This problem highlights advanced calculus concepts suitable for university-level students.

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