The given integral is:

$\int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx$

To solve this, we can apply the substitution method. Let's set:

$u = \sqrt{x} \quad \text{so that} \quad x = u^2 \quad \text{and} \quad dx = 2u \, du$

Now substitute $u$ and $dx$ into the integral:

$\int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx = \int \frac{e^u}{u} \cdot 2u \, du = 2 \int e^u \, du$

The integral of $e^u$ is simply $e^u$, so we have:

$2 \int e^u \, du = 2e^u + C$

Finally, substitute back $u = \sqrt{x}$ to get the solution in terms of $x$:

$2e^u + C = 2e^{\sqrt{x}} + C$

Thus, the solution to the integral is:

$\boxed{2e^{\sqrt{x}} + C}$

Would you like further details or clarification?

Here are 5 related questions to explore further:

1. How would you solve $\int e^{x^2} \, dx$?
2. What is the significance of substitution in integrals involving square roots?
3. Can you derive a similar solution for $\int e^{x^{1/3}} \, dx$?
4. What happens if the exponential is in the denominator, e.g. $\int \frac{1}{e^{\sqrt{x}}} \, dx$?
5. How would you handle definite integrals with limits for this problem?

Tip: When dealing with integrals involving roots or powers, substitution often simplifies the process greatly by reducing the expression to a more familiar form.