Evaluate: `int_(1)^(2) (sqrt(x))/(sqrt(3-x) + sqrt(x)) dx`.

Answer

To evaluate the integral \( I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x} + \sqrt{x}} \, dx \), we can use a symmetry property of definite integrals.

### Step-by-step Solution:

1. **Define the Integral:**
Let \( I = \int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x} + \sqrt{x}} \, dx \).

2. **Use the Symmetry Property:**
We can use the property of integrals that states:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
Here, \( a = 1 \) and \( b = 2 \), so \( a + b = 3 \). Thus, we have:
\[ I = \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{x} + \sqrt{3-x}} \, dx \]

3. **Rewrite the Integral:**
Now we can express this integral as:
\[ I = \int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{x} + \sqrt{3-x}} \, dx \]

4. **Add the Two Integrals:**
Now, we can add the two expressions for \( I \):
\[ 2I = \int_{1}^{2} \left( \frac{\sqrt{x}}{\sqrt{3-x} + \sqrt{x}} + \frac{\sqrt{3-x}}{\sqrt{x} + \sqrt{3-x}} \right) dx \]

5. **Simplify the Expression:**
Notice that:
\[ \frac{\sqrt{x}}{\sqrt{3-x} + \sqrt{x}} + \frac{\sqrt{3-x}}{\sqrt{x} + \sqrt{3-x}} = 1 \]
Therefore:
\[ 2I = \int_{1}^{2} 1 \, dx \]

6. **Evaluate the Integral:**
Now we can evaluate the integral:
\[ 2I = \int_{1}^{2} 1 \, dx = [x]_{1}^{2} = 2 - 1 = 1 \]

7. **Solve for \( I \):**
Thus, we have:
\[ 2I = 1 \implies I = \frac{1}{2} \]

### Final Answer:
The value of the integral is:
\[ I = \frac{1}{2} \]