The given improper integral is:

$\int_0^\infty \frac{x}{x^3 + 1} \, dx$

We are tasked with determining whether the integral converges or diverges using the Comparison Theorem.

Step-by-step approach:

1. Behavior of the integrand: The comparison theorem requires us to compare the integrand $\frac{x}{x^3 + 1}$ with a simpler function that we know either converges or diverges.

   • For small $x$ (near $x = 0$):
     When $x \to 0$, $\frac{x}{x^3 + 1} \approx \frac{x}{1} = x$.
     The integral $\int_0^1 x \, dx$ converges.

   • For large $x$ (as $x \to \infty$):
     When $x$ is large, $x^3$ dominates $1$ in the denominator, so the integrand behaves like: $\frac{x}{x^3 + 1} \approx \frac{x}{x^3} = \frac{1}{x^2}$ The integral $\int_1^\infty \frac{1}{x^2} \, dx$ converges.

2. Applying the Comparison Theorem:

   • For large $x$, compare $\frac{x}{x^3 + 1}$ with $\frac{1}{x^2}$.
   • We know that $\int_1^\infty \frac{1}{x^2} \, dx$ converges because it is a $p$-integral with $p = 2 > 1$.

3. Conclusion:
   Since the integrand behaves similarly to $\frac{1}{x^2}$ for large $x$, and $\frac{x}{x^3 + 1}$ is smaller than $\frac{1}{x^2}$ for sufficiently large $x$, by the Comparison Theorem, the given integral converges.

Would you like more details on any step, or do you have any questions?

Related questions:

1. How do we determine the convergence of improper integrals using the Limit Comparison Test?
2. What are some common comparison functions used in the Comparison Theorem?
3. Why does the $p$-integral $\int_1^\infty \frac{1}{x^p} dx$ converge when $p > 1$?
4. How would the integral change if the limits were different (e.g., $\int_1^\infty \frac{x}{x^3 + 1} dx$)?
5. Can the Comparison Theorem be used with divergent integrals as well?

Tip:

For large values of $x$, identifying the dominant term in the denominator or numerator helps simplify the behavior of functions when applying the Comparison Theorem.