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 the integrand as follows:

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

This simplifies to:

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

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

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

### Step 3: Integrate term by term
Now we can integrate each term separately:

1. For \( x^{\frac{5}{2}} \):

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

2. For \( 3x^{\frac{3}{2}} \):

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

3. For \( 4x^{-\frac{1}{2}} \):

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

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

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

where \( C \) is the constant of integration.

### Final Answer
Thus, the final answer is:

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

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