The given problem asks to evaluate the indefinite integral:

$\int \left( \frac{3\sqrt{x} + 2}{\sqrt{x}} \right) dx$

Step-by-step Solution:

1. Simplify the integrand: Break the integrand into two separate terms:

   $\frac{3\sqrt{x}}{\sqrt{x}} + \frac{2}{\sqrt{x}} = 3 + \frac{2}{\sqrt{x}}$

   Since $\frac{3\sqrt{x}}{\sqrt{x}} = 3$ and $\frac{2}{\sqrt{x}}$ can be rewritten as $2x^{-\frac{1}{2}}$, the expression simplifies to:

   $3 + 2x^{-\frac{1}{2}}$

2. Integrate term by term: Now, we can integrate each term separately.

   • The integral of $3$ is:

     $\int 3 dx = 3x$

   • The integral of $2x^{-\frac{1}{2}}$ is:

     $\int 2x^{-\frac{1}{2}} dx = 2 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 4x^{\frac{1}{2}}$

3. Final answer: Combining both results:

   $3x + 4\sqrt{x} + C$

   Since the problem specifies not to write the arbitrary constant $C$, the final result is:

   $3x + 4\sqrt{x}$

Would you like more details or clarifications on any part of the solution?

Here are 5 related questions for further practice:

1. How would you evaluate $\int \frac{1}{\sqrt{x}} dx$?
2. What is the indefinite integral of $x^n$ for any real $n$?
3. How would you find $\int \left( 5x^{3} + 2x^{-1} \right) dx$?
4. How do you apply the power rule of integration to negative exponents?
5. What would change in the evaluation if the limits of integration were from 1 to 4?

Tip: When integrating, always simplify the integrand as much as possible before applying basic rules.