Evaluate : `int_(0)^(9) (sqrtx)/(sqrtx+sqrt(9-x)) dx`

Answer

To evaluate the integral

\[ I = \int_{0}^{9} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{9 - x}} \, dx, \]

we can use a property of definite integrals. We will first find a second expression for the integral by substituting \( x \) with \( 9 - x \).

### Step 1: Substitute \( x \) with \( 9 - x \)

Let \( u = 9 - x \). Then, \( dx = -du \). The limits change as follows:
- When \( x = 0 \), \( u = 9 \)
- When \( x = 9 \), \( u = 0 \)

Thus, we can rewrite the integral as:

\[ I = \int_{9}^{0} \frac{\sqrt{9 - u}}{\sqrt{9 - u} + \sqrt{u}} (-du) = \int_{0}^{9} \frac{\sqrt{9 - u}}{\sqrt{9 - u} + \sqrt{u}} \, du. \]

### Step 2: Rewrite the integral

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

1.
\[ I = \int_{0}^{9} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{9 - x}} \, dx \]

2.
\[ I = \int_{0}^{9} \frac{\sqrt{9 - x}}{\sqrt{9 - x} + \sqrt{x}} \, dx. \]

### Step 3: Add the two expressions for \( I \)

Now, let's add both expressions for \( I \):

\[ 2I = \int_{0}^{9} \left( \frac{\sqrt{x}}{\sqrt{x} + \sqrt{9 - x}} + \frac{\sqrt{9 - x}}{\sqrt{9 - x} + \sqrt{x}} \right) dx. \]

### Step 4: Simplify the integrand

Notice that the two fractions can be combined:

\[ \frac{\sqrt{x}}{\sqrt{x} + \sqrt{9 - x}} + \frac{\sqrt{9 - x}}{\sqrt{9 - x} + \sqrt{x}} = \frac{\sqrt{x}(\sqrt{9 - x} + \sqrt{x}) + \sqrt{9 - x}(\sqrt{x} + \sqrt{9 - x})}{(\sqrt{x} + \sqrt{9 - x})(\sqrt{9 - x} + \sqrt{x})}. \]

This simplifies to:

\[ \frac{\sqrt{x} + \sqrt{9 - x}}{\sqrt{x} + \sqrt{9 - x}} = 1. \]

### Step 5: Evaluate the integral

Thus, we have:

\[ 2I = \int_{0}^{9} 1 \, dx = [x]_{0}^{9} = 9 - 0 = 9. \]

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

Now, we can solve for \( I \):

\[ 2I = 9 \implies I = \frac{9}{2}. \]

### Final Answer

Thus, the value of the integral is

\[ \boxed{\frac{9}{2}}. \]