The integral you're asking to evaluate is:

$\int_{-\infty}^{\infty} \frac{\sqrt{e^{x^2}}}{e^{x^2}} \, dx$

First, let's simplify the integrand:

$\frac{\sqrt{e^{x^2}}}{e^{x^2}} = \frac{e^{x^2/2}}{e^{x^2}} = e^{-x^2/2}$

So, the integral becomes:

$\int_{-\infty}^{\infty} e^{-x^2/2} \, dx$

This is a standard Gaussian integral with a coefficient in the exponent. The general formula for the Gaussian integral is:

$\int_{-\infty}^{\infty} e^{-ax^2} \, dx = \sqrt{\frac{\pi}{a}}$

Here, $a = \frac{1}{2}$. Therefore, the integral evaluates to:

$\int_{-\infty}^{\infty} e^{-x^2/2} \, dx = \sqrt{\frac{2\pi}{1}} = \sqrt{2\pi}$

Thus, the final answer is:

$\sqrt{2\pi}$

Would you like further details or have any questions? Here are some related questions to expand your understanding:

1. How do Gaussian integrals relate to probability theory?
2. What is the significance of the standard normal distribution in statistics?
3. How can we evaluate integrals involving $e^{-x^2}$ using polar coordinates?
4. How would the integral change if the exponent were $e^{-ax^2}$ with $a \neq \frac{1}{2}$?
5. What are the applications of Gaussian functions in physics?

Tip: When dealing with integrals involving exponentials, always look for a Gaussian form, as these integrals have well-known solutions.