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The given problem asks whether the improper integral

\[
\int_0^{\infty} \frac{dx}{\sqrt{x^2 + 1}}
\]

converges or diverges.

### Solution:

1. **Integral analysis**:
   The integrand is \( \frac{1}{\sqrt{x^2 + 1}} \), and we are tasked with evaluating the integral over the interval \([0, \infty)\). This is an improper integral due to the infinite upper bound, so we must analyze its convergence.

2. **Behavior at \( x = 0 \)**:
   At \(x = 0\), the integrand becomes:

   \[
   \frac{1}{\sqrt{0^2 + 1}} = 1
   \]

   So near \( x = 0 \), the function is bounded and well-behaved. There are no issues with convergence at the lower limit.

3. **Behavior as \( x \to \infty \)**:
   For large \(x\), we can approximate \( \sqrt{x^2 + 1} \approx x \), so the integrand behaves like:

   \[
   \frac{1}{\sqrt{x^2 + 1}} \approx \frac{1}{x}
   \]

   The integral of \( \frac{1}{x} \) from some large number to infinity is:

   \[
   \int \frac{1}{x} \, dx = \ln(x)
   \]

   Since \( \ln(x) \) diverges as \( x \to \infty \), we expect that the integral might converge or diverge depending on the specific form of the function.

4. **Exact evaluation**:
   To evaluate the integral exactly, perform the following substitution:

   Let \( x = \sinh(t) \), where \( \sinh(t) = \frac{e^t - e^{-t}}{2} \), then we know that:

   \[
   \frac{dx}{\sqrt{x^2 + 1}} = \frac{1}{\cosh(t)} \, dt
   \]

   Therefore, the integral simplifies to a well-known convergent form:

   \[
   \int_0^{\infty} \frac{dt}{\cosh(t)} = \frac{\pi}{2}
   \]

### Conclusion:

The integral converges, and its exact value is \( \frac{\pi}{2} \).

Would you like any further explanation or details?

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Here are 5 related questions for further exploration:

1. How do you determine if an improper integral converges?
2. What substitution methods are useful in evaluating integrals like this?
3. How does the behavior of a function at infinity affect the convergence of an integral?
4. What other types of improper integrals are commonly encountered?
5. Can you apply the same technique to integrals with different powers of \(x^2 + 1\)?

**Tip**: When dealing with improper integrals, always check the behavior of the function at both the lower and upper limits of the integration range.

Определите, сходится или расходится интеграл: ∫(dx / √(x² + 1)) от 0 до ∞

This problem explores whether the improper integral ∫(dx / √(x² + 1)) from 0 to ∞ converges or diverges. It covers key calculus concepts such as improper integrals, convergence behavior, and hyperbolic substitutions for solving the integral.

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