`int_(1)^(7)(log sqrt(x))/(log sqrt(8-x)+log sqrt(x))dx=`

Answer

To solve the integral

\[ I = \int_{1}^{7} \frac{\log \sqrt{x}}{\log \sqrt{8-x} + \log \sqrt{x}} \, dx, \]

we can use a property of definite integrals. This property states that if we have an integral of the form

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx, \]

we can simplify our calculations. Here, \(a = 1\) and \(b = 7\), so \(a + b = 8\).

### Step 1: Apply the property

We can rewrite the integral \(I\) as follows:

\[ I = \int_{1}^{7} \frac{\log \sqrt{8-x}}{\log \sqrt{x} + \log \sqrt{8-x}} \, dx. \]

### Step 2: Combine the two integrals

Now we have two expressions for \(I\):

1. \(I = \int_{1}^{7} \frac{\log \sqrt{x}}{\log \sqrt{8-x} + \log \sqrt{x}} \, dx\)
2. \(I = \int_{1}^{7} \frac{\log \sqrt{8-x}}{\log \sqrt{x} + \log \sqrt{8-x}} \, dx\)

Let's add these two equations:

\[ 2I = \int_{1}^{7} \left( \frac{\log \sqrt{x}}{\log \sqrt{8-x} + \log \sqrt{x}} + \frac{\log \sqrt{8-x}}{\log \sqrt{x} + \log \sqrt{8-x}} \right) dx. \]

### Step 3: Simplify the integrand

Notice that the two fractions in the integrand add up to 1:

\[ \frac{\log \sqrt{x}}{\log \sqrt{8-x} + \log \sqrt{x}} + \frac{\log \sqrt{8-x}}{\log \sqrt{x} + \log \sqrt{8-x}} = 1. \]

So we have:

\[ 2I = \int_{1}^{7} 1 \, dx. \]

### Step 4: Evaluate the integral

Now we can evaluate the integral:

\[ \int_{1}^{7} 1 \, dx = 7 - 1 = 6. \]

### Step 5: Solve for \(I\)

Thus, we have:

\[ 2I = 6 \implies I = \frac{6}{2} = 3. \]

### Final Answer

The value of the integral is

\[ \boxed{3}. \]