Evaluate the following integrals.
`int (x^(3) + 3x^(2) + 4)/(sqrt(x)) dx`

Answer

To evaluate the integral

\[ \int \frac{x^3 + 3x^2 + 4}{\sqrt{x}} \, dx, \]

we can start by simplifying the integrand.

### Step 1: Rewrite the integrand
We can express \(\sqrt{x}\) as \(x^{1/2}\). Thus, we can rewrite the integrand as:

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

This simplifies to:

\[ x^{5/2} + 3x^{3/2} + 4x^{-1/2}. \]

### Step 2: Set up the integral
Now we can rewrite the integral as:

\[ \int \left( x^{5/2} + 3x^{3/2} + 4x^{-1/2} \right) \, dx. \]

### Step 3: Integrate term by term
We will integrate each term separately using the power rule for integration, which states that:

\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1). \]

1. For the first term \(x^{5/2}\):

\[ \int x^{5/2} \, dx = \frac{x^{5/2 + 1}}{5/2 + 1} = \frac{x^{7/2}}{7/2} = \frac{2}{7} x^{7/2}. \]

2. For the second term \(3x^{3/2}\):

\[ \int 3x^{3/2} \, dx = 3 \cdot \frac{x^{3/2 + 1}}{3/2 + 1} = 3 \cdot \frac{x^{5/2}}{5/2} = \frac{6}{5} x^{5/2}. \]

3. For the third term \(4x^{-1/2}\):

\[ \int 4x^{-1/2} \, dx = 4 \cdot \frac{x^{-1/2 + 1}}{-1/2 + 1} = 4 \cdot \frac{x^{1/2}}{1/2} = 8x^{1/2}. \]

### Step 4: Combine the results
Now we can combine all the results:

\[ \int \frac{x^3 + 3x^2 + 4}{\sqrt{x}} \, dx = \frac{2}{7} x^{7/2} + \frac{6}{5} x^{5/2} + 8x^{1/2} + C. \]

### Final Answer
Thus, the final result of the integral is:

\[ \int \frac{x^3 + 3x^2 + 4}{\sqrt{x}} \, dx = \frac{2}{7} x^{7/2} + \frac{6}{5} x^{5/2} + 8x^{1/2} + C. \]