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The problem is asking for the evaluation of the definite integral:

\[
\int_{1}^{4} \left( 2x + \frac{3}{\sqrt{x}} \right) \, dx
\]

Let's solve this step by step.

### Step 1: Break down the integral
The integral can be split into two separate integrals:

\[
\int_{1}^{4} \left( 2x + \frac{3}{\sqrt{x}} \right) \, dx = \int_{1}^{4} 2x \, dx + \int_{1}^{4} \frac{3}{\sqrt{x}} \, dx
\]

### Step 2: Evaluate each integral

#### Integral 1: \(\int 2x \, dx\)
The integral of \(2x\) with respect to \(x\) is:

\[
\int 2x \, dx = x^2 + C
\]

So,

\[
\int_{1}^{4} 2x \, dx = \left[ x^2 \right]_{1}^{4} = (4^2) - (1^2) = 16 - 1 = 15
\]

#### Integral 2: \(\int \frac{3}{\sqrt{x}} \, dx\)
We can rewrite \(\frac{3}{\sqrt{x}}\) as \(3x^{-\frac{1}{2}}\). The integral of \(3x^{-\frac{1}{2}}\) is:

\[
\int 3x^{-\frac{1}{2}} \, dx = 3 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 6\sqrt{x} + C
\]

So,

\[
\int_{1}^{4} \frac{3}{\sqrt{x}} \, dx = \left[ 6\sqrt{x} \right]_{1}^{4} = 6\sqrt{4} - 6\sqrt{1} = 6(2) - 6(1) = 12 - 6 = 6
\]

### Step 3: Add the results
Now, add the results of the two integrals:

\[
\int_{1}^{4} \left( 2x + \frac{3}{\sqrt{x}} \right) \, dx = 15 + 6 = 21
\]

### Final Answer:
The value of the definite integral is \( \boxed{21} \).

Would you like to go over any part of this solution in more detail or have any additional questions?

Here are 5 related questions:
1. How would you integrate \(2x\) using substitution?
2. What is the general formula for integrating \(x^n\)?
3. How does the Fundamental Theorem of Calculus apply to definite integrals?
4. What would the result be if the limits of integration were reversed?
5. How can integration by parts be applied to solve integrals involving products of functions?

**Tip:** When solving definite integrals, always remember to evaluate at the upper and lower limits and subtract to find the final value.

Evaluate the definite integral from 1 to 4 of (2x + 3/sqrt(x)) dx.

Learn how to evaluate the definite integral of (2x + 3/sqrt(x)) from 1 to 4. This problem covers key calculus concepts such as definite integrals and the Fundamental Theorem of Calculus. Suitable for students in Grades 11-12.

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