1. **State the problem:** We want to determine whether the series $$\sum_{n=2}^{\infty} \frac{n^2}{n^3 + 1}$$ converges or diverges using the integral test. 2. **Set up the integral test:** Consider the function $$f(x) = \frac{x^2}{x^3 + 1}$$ which is positive, continuous, and decreasing for $$x \geq 2$$. 3. **Evaluate the improper integral:** $$\int_2^{\infty} \frac{x^2}{x^3 + 1} \, dx$$ 4. **Use substitution:** Let $$u = x^3 + 1$$, then $$du = 3x^2 dx$$ or $$x^2 dx = \frac{du}{3}$$. 5. **Change the limits:** When $$x=2$$, $$u = 2^3 + 1 = 9$$; when $$x \to \infty$$, $$u \to \infty$$. 6. **Rewrite the integral:** $$\int_9^{\infty} \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3} \int_9^{\infty} \frac{1}{u} du$$ 7. **Integrate:** $$\frac{1}{3} [\ln|u|]_9^{\infty} = \frac{1}{3} (\lim_{t \to \infty} \ln t - \ln 9)$$ 8. **Evaluate the limit:** $$\lim_{t \to \infty} \ln t = \infty$$, so the integral diverges. 9. **Conclusion:** Since the integral diverges, by the integral test, the series $$\sum_{n=2}^{\infty} \frac{n^2}{n^3 + 1}$$ also diverges. **Final answer:** The series diverges. DIVERGES ❌