To solve this question, we analyze the improper integral $I = \int_{0}^\infty \frac{2e^{x} + 3x}{3e^{2x} + 3x} \, dx$ and compare it with the simpler integrals $J$, $K$, and $L$ provided.

Step-by-Step Analysis:

1. Simplify $J = \int_{0}^\infty \frac{5e^x}{3e^{2x} + 3x} \, dx$:

   • For large $x$, $3e^{2x}$ dominates $3x$ in the denominator, so the fraction $\frac{5e^x}{3e^{2x} + 3x}$ behaves like $\frac{5e^x}{3e^{2x}} = \frac{5}{3e^x}$, which decays to $0$.
   • $J$ converges because the terms decay quickly as $x \to \infty$.

2. Simplify $L = \int_{0}^\infty \frac{5e^x}{3x} \, dx$:

   • For large $x$, the term $\frac{5e^x}{3x}$ grows very rapidly due to $e^x$, leading to divergence. Therefore, $L = \infty$.

3. Analyze $I$:

   • $I$ involves a more complex numerator $2e^x + 3x$ and denominator $3e^{2x} + 3x$, but for large $x$, the dominant term is $\frac{2e^x}{3e^{2x}} = \frac{2}{3e^x}$, which suggests convergence.
   • However, by the comparison test:
     • Since $J \leq I$ and $L \geq I$, and $L$ diverges ($\infty$), $I$ also diverges.

4. Validate with $K$:

   • $K = \int_{0}^\infty \frac{5}{3e^x} \, dx$ simplifies to $\int_{0}^\infty \frac{5}{3} e^{-x} \, dx$, which converges because $e^{-x}$ decays rapidly as $x \to \infty$.

Final Answer:

The correct statement is: $\mathbf{I \geq J \geq L = \infty, \text{ hence } I \text{ is divergent.}}$

Any questions or areas you'd like clarified?

Here are five follow-up questions to expand on this explanation:

1. How does the comparison test determine the convergence or divergence of improper integrals?
2. Why does $\int_{0}^\infty e^{-x} \, dx$ converge, while $\int_{0}^\infty e^x \, dx$ diverges?
3. Can you generalize the behavior of exponential terms $e^{kx}$ (for $k > 0$) in integrals as $x \to \infty$?
4. What other convergence tests could be applied to this integral $I$, such as the limit comparison test or ratio test?
5. How do small terms in the denominator (e.g., $3x$ compared to $3e^{2x}$) affect the convergence/divergence of an integral?

Tip: When solving improper integrals, always analyze the dominant terms in both the numerator and denominator for large $x$, as they dictate the integral's behavior.