The value of the integral `int_3^6 sqrtx/(sqrt(9-x)+sqrtx)dx` is

Explanation

To evaluate the integral

\[ I = \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx, \]

we can use the property of definite integrals that states:

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

### Step 1: Apply the property of definite integrals

Let’s apply this property to our integral. Here, \( a = 3 \) and \( b = 6 \), so \( b + a = 9 \). We can rewrite the integral as:

\[ I = \int_{3}^{6} \frac{\sqrt{9 - x}}{\sqrt{x} + \sqrt{9 - x}} \, dx. \]

### Step 2: Rewrite the integral

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

1. \( I = \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx \) (Equation 1)
2. \( I = \int_{3}^{6} \frac{\sqrt{9 - x}}{\sqrt{x} + \sqrt{9 - x}} \, dx \) (Equation 2)

### Step 3: Add the two equations

Now, we can add both equations:

\[ 2I = \int_{3}^{6} \left( \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} + \frac{\sqrt{9 - x}}{\sqrt{x} + \sqrt{9 - x}} \right) \, dx. \]

### Step 4: Simplify the expression

Notice that the denominators are the same, so we can combine the fractions:

\[ 2I = \int_{3}^{6} \frac{\sqrt{x} + \sqrt{9 - x}}{\sqrt{9-x} + \sqrt{x}} \, dx. \]

This simplifies to:

\[ 2I = \int_{3}^{6} 1 \, dx. \]

### Step 5: Evaluate the integral

Now we can evaluate the integral:

\[ 2I = \int_{3}^{6} 1 \, dx = [x]_{3}^{6} = 6 - 3 = 3. \]

### Step 6: Solve for \( I \)

Now, we can find \( I \):

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

Thus, the value of the integral is

\[ \boxed{\frac{3}{2}}. \]