The problem is to evaluate the indefinite integral of the function $\sqrt{x} - \frac{1}{2}x + \frac{2}{\sqrt{x}}$ with respect to $x$. In other words, we want to find $\int (\sqrt{x} - \frac{1}{2}x + \frac{2}{\sqrt{x}}) dx$.

1. Problem Description

The problem is to evaluate the indefinite integral of the function $\sqrt{x} - \frac{1}{2}x + \frac{2}{\sqrt{x}}$ with respect to $x$. In other words, we want to find $\int (\sqrt{x} - \frac{1}{2}x + \frac{2}{\sqrt{x}}) dx$.

2. Solution Steps

First, rewrite the terms in the integrand using exponents:

$\int (\sqrt{x} - \frac{1}{2}x + \frac{2}{\sqrt{x}}) dx = \int (x^{1/2} - \frac{1}{2}x + 2x^{-1/2}) dx$.

Now, we apply the power rule for integration to each term. The power rule states that $\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$.

Applying the power rule to each term, we get:

$\int x^{1/2} dx = \frac{x^{1/2 + 1}}{1/2 + 1} = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$

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

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

Adding the results and the constant of integration $C$, we have:

$\int (x^{1/2} - \frac{1}{2}x + 2x^{-1/2}) dx = \frac{2}{3}x^{3/2} - \frac{1}{4}x^2 + 4x^{1/2} + C$

We can rewrite this in terms of square roots:

$\frac{2}{3}x^{3/2} - \frac{1}{4}x^2 + 4x^{1/2} + C = \frac{2}{3}x\sqrt{x} - \frac{1}{4}x^2 + 4\sqrt{x} + C$.

3. Final Answer

$\frac{2}{3}x\sqrt{x} - \frac{1}{4}x^2 + 4\sqrt{x} + C$