To determine whether the given improper integral is convergent or divergent, we'll apply the Comparison Theorem. The integral in question is:

$\int_0^\infty \frac{13x}{x^3 + 1} \, dx$

Step 1: Analyzing the Behavior of the Integrand

The Comparison Theorem is helpful when comparing the given function to simpler functions whose integrals we already know to be convergent or divergent. Let's first examine the behavior of the integrand for large $x$ (as $x \to \infty$) and near $x = 0$.

For large values of $x$, the denominator $x^3 + 1$ behaves like $x^3$. Thus, for large $x$:

$\frac{13x}{x^3 + 1} \approx \frac{13x}{x^3} = \frac{13}{x^2}$

We know that the integral $\int_1^\infty \frac{1}{x^2} \, dx$ converges because:

$\int_1^\infty \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_1^\infty = 1$

So, for large $x$, the integrand behaves similarly to $\frac{13}{x^2}$, and we can compare the given function to $\frac{13}{x^2}$ to examine its convergence at infinity.

Step 2: Applying the Comparison Theorem

To apply the Comparison Theorem:

• For large $x$, we know $\frac{13x}{x^3 + 1} \leq \frac{13}{x^2}$, because $x^3 + 1 \geq x^3$ for all $x \geq 1$.
• The integral $\int_1^\infty \frac{13}{x^2} \, dx$ converges.
• Since $\frac{13x}{x^3 + 1}$ is dominated by $\frac{13}{x^2}$, the integral $\int_1^\infty \frac{13x}{x^3 + 1} \, dx$ also converges by the Comparison Theorem.

Step 3: Behavior Near $x = 0$

Near $x = 0$, the integrand behaves like $\frac{13x}{x^3 + 1} \approx 13x$ because $x^3 + 1 \approx 1$ when $x$ is small. The integral:

$\int_0^\epsilon 13x \, dx$

for some small $\epsilon > 0$, is clearly convergent since:

$\int_0^\epsilon 13x \, dx = \frac{13x^2}{2} \Big|_0^\epsilon = \frac{13\epsilon^2}{2}$

Thus, the integral is also convergent near $x = 0$.

Conclusion:

By the Comparison Theorem, the given integral is convergent. The integrand behaves similarly to $\frac{13}{x^2}$ as $x \to \infty$, and the function is well-behaved near $x = 0$.

---

Would you like more details on any of these steps or have additional questions? Here are some related questions you might find interesting:

1. How can you apply the Comparison Theorem to other types of improper integrals?
2. What would happen if the denominator were $x^2 + 1$ instead of $x^3 + 1$?
3. How does the behavior of an integrand near infinity affect convergence?
4. Can we use the Limit Comparison Test for this problem?
5. What are some other common tests for convergence of improper integrals?

Tip: Always check the behavior of the integrand both at the lower limit and as $x \to \infty$ to decide how to compare with simpler, known integrals.