`int (x^(2) +2x -5)/(sqrt(x))dx`

Answer

To solve the integral \( \int \frac{x^2 + 2x - 5}{\sqrt{x}} \, dx \), we can start by separating the terms in the numerator.

### Step 1: Separate the terms
We can rewrite the integral as:
\[ \int \frac{x^2}{\sqrt{x}} \, dx + \int \frac{2x}{\sqrt{x}} \, dx - \int \frac{5}{\sqrt{x}} \, dx \]

### Step 2: Simplify each term
Now we simplify each term:
- The first term becomes \( \int x^{2 - \frac{1}{2}} \, dx = \int x^{\frac{3}{2}} \, dx \)
- The second term becomes \( \int 2x^{1 - \frac{1}{2}} \, dx = \int 2x^{\frac{1}{2}} \, dx \)
- The third term becomes \( \int 5x^{-\frac{1}{2}} \, dx \)

So, we can rewrite the integral as:
\[ \int x^{\frac{3}{2}} \, dx + 2 \int x^{\frac{1}{2}} \, dx - 5 \int x^{-\frac{1}{2}} \, dx \]

### Step 3: Integrate each term
Now we integrate each term:
1. For \( \int x^{\frac{3}{2}} \, dx \):
\[ = \frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1} = \frac{x^{\frac{5}{2}}}{\frac{5}{2}} = \frac{2}{5} x^{\frac{5}{2}} \]

2. For \( 2 \int x^{\frac{1}{2}} \, dx \):
\[ = 2 \cdot \frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} = 2 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}} = \frac{4}{3} x^{\frac{3}{2}} \]

3. For \( -5 \int x^{-\frac{1}{2}} \, dx \):
\[ = -5 \cdot \frac{x^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} = -5 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = -10 x^{\frac{1}{2}} \]

### Step 4: Combine the results
Now we combine all the results:
\[ \int \frac{x^2 + 2x - 5}{\sqrt{x}} \, dx = \frac{2}{5} x^{\frac{5}{2}} + \frac{4}{3} x^{\frac{3}{2}} - 10 x^{\frac{1}{2}} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{x^2 + 2x - 5}{\sqrt{x}} \, dx = \frac{2}{5} x^{\frac{5}{2}} + \frac{4}{3} x^{\frac{3}{2}} - 10 x^{\frac{1}{2}} + C \]