To determine the convergence of the given improper integral:

$A = \int_{1}^{\infty} \frac{2x^2 + x + 1}{e^{4x} + x^3 + 3} dx,$

we analyze the integrand $\frac{2x^2 + x + 1}{e^{4x} + x^3 + 3}$ as $x \to \infty$. Here are the steps:

Step 1: Dominance of terms in the denominator for large $x$

The denominator $e^{4x} + x^3 + 3$ grows very quickly as $x \to \infty$ because $e^{4x}$ dominates all other terms. Therefore, asymptotically: $e^{4x} + x^3 + 3 \sim e^{4x} \quad \text{as } x \to \infty.$

Step 2: Simplify the integrand for large $x$

For large $x$, the integrand approximates: $\frac{2x^2 + x + 1}{e^{4x} + x^3 + 3} \sim \frac{2x^2 + x + 1}{e^{4x}}.$

Step 3: Analyze the asymptotic behavior of the simplified form

The simplified form becomes: $\frac{2x^2 + x + 1}{e^{4x}} = \frac{2x^2}{e^{4x}} + \frac{x}{e^{4x}} + \frac{1}{e^{4x}}.$ Each term in this sum decays exponentially fast as $x \to \infty$, since $e^{4x}$ in the denominator grows much faster than any polynomial term in the numerator. Hence, the integrand tends to zero extremely quickly.

Step 4: Check convergence near $x = 1$

For $x \geq 1$, the denominator $e^{4x} + x^3 + 3$ is strictly positive and the numerator $2x^2 + x + 1$ is a polynomial. The integrand is therefore well-defined and finite for $x \geq 1$.

Step 5: Conclude convergence

Since the integrand decays exponentially fast as $x \to \infty$, and it is well-defined and finite for $x \geq 1$, the improper integral converges.

$\boxed{\text{The integral converges.}}$

Would you like me to compute the exact value or explain another aspect in more detail?

Here are 5 related questions:

1. How does exponential decay affect the convergence of integrals?
2. What role does the denominator's growth rate play in integral convergence?
3. How would the convergence change if the exponential term $e^{4x}$ were replaced with $x^4$?
4. Can the comparison test be used to formally prove convergence here?
5. How does polynomial growth in the numerator compare to other functions for integral behavior?

Tip: When analyzing improper integrals, always compare the growth rates of the numerator and denominator for large $x$.