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To determine whether the given integral 

\[
\int_0^\infty \frac{e^x}{e^{\alpha x} + x^2} \, dx
\]

converges when \(\alpha > 1\), we will use the **comparison test**.

### Step 1: Analyze the behavior of the integrand for large \(x\)
- For large \(x\), the term \(e^{\alpha x}\) in the denominator dominates \(x^2\) because exponential growth is much faster than polynomial growth. 
- Therefore, for large \(x\), the integrand behaves approximately like:
  \[
  \frac{e^x}{e^{\alpha x}} = e^{(1-\alpha)x}.
  \]

Since \(\alpha > 1\), the exponent \(1-\alpha\) is negative, and \(e^{(1-\alpha)x}\) decays exponentially as \(x \to \infty\). The integral of this type of function converges.

### Step 2: Analyze the behavior of the integrand for small \(x\)
- For small \(x\), the denominator \(e^{\alpha x} + x^2 \approx 1 + x^2\) (since \(e^{\alpha x} \approx 1\) for small \(x\)).
- The integrand behaves like:
  \[
  \frac{e^x}{1 + x^2}.
  \]
This function is finite and integrable over the range \(0 \leq x \leq 1\).

### Step 3: Use a formal comparison test
We can split the integral into two regions:
1. \(x \in [0, 1]\): The integrand is finite and bounded because \(\frac{e^x}{e^{\alpha x} + x^2} \leq \frac{e^x}{1 + x^2}\), which is integrable over this range.
2. \(x \in [1, \infty)\): For large \(x\), we compare \(\frac{e^x}{e^{\alpha x} + x^2}\) with \(e^{(1-\alpha)x}\). Since \(e^{(1-\alpha)x}\) is integrable over \([1, \infty)\) when \(\alpha > 1\), the original integrand is smaller than or comparable to an integrable function.

Thus, by the comparison test, the given integral converges when \(\alpha > 1\).

Would you like me to elaborate on any part of this solution? Here are five related questions to explore:

1. How does the exponential growth rate compare to polynomial growth?
2. What is the comparison test for improper integrals?
3. Why does \(e^{(1-\alpha)x}\) lead to convergence when \(\alpha > 1\)?
4. How does the behavior of the integrand near \(x = 0\) affect convergence?
5. Can this integral be evaluated explicitly, or is only convergence important?

**Tip:** Always analyze the asymptotic behavior (both for large and small \(x\)) when testing for the convergence of improper integr

Does the integral \( \int_0^\infty \frac{e^x}{e^{\alpha x} + x^2} \, dx \) converge when \(\alpha > 1\)?

Learn how to determine the convergence of the improper integral \( \int_0^\infty \frac{e^x}{e^{\alpha x} + x^2} \, dx \) when \( \alpha > 1 \) using the comparison test. This problem involves exponential functions, asymptotic behavior for large \(x\), and integration techniques suitable for advanced high school or college-level students.

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