`int_(a)^(b)(sqrt(x))/(sqrt(x)+sqrt(a)+b-x) =dx=......`

Options

A. `a+b`
B. `(b-a)/(2)`
C. `a-b`
D. `(a-b)/(2)`

Explanation

To solve the integral

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

we can use a property of definite integrals. The property states that:

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

### Step 1: Apply the property

We will apply this property to our integral. First, we need to express \( f(a + b - x) \):

\[ f(a + b - x) = \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{a} + b - (a + b - x)}. \]

Simplifying the denominator:

\[ b - (a + b - x) = x - a, \]

thus,

\[ f(a + b - x) = \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{a} + x - a}. \]

### Step 2: Set up the new integral

Now we have:

\[ I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{a} + x - a} \, dx. \]

### Step 3: Add the two integrals

We now have two expressions for \( I \):

1. \( I = \int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a} + b - x} \, dx \)
2. \( I = \int_{a}^{b} \frac{\sqrt{a + b - x}}{\sqrt{a + b - x} + \sqrt{a} + x - a} \, dx \)

Adding these two integrals gives:

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

### Step 4: Simplify the integrand

The common denominator for both fractions is:

\[ \sqrt{x} + \sqrt{a} + b - x. \]

Thus, we can combine the numerators:

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

### Step 5: Evaluate the integral

Now we evaluate:

\[ \int_{a}^{b} 1 \, dx = b - a. \]

So we have:

\[ 2I = b - a. \]

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

Dividing both sides by 2 gives:

\[ I = \frac{b - a}{2}. \]

### Final Answer

Thus, the value of the integral is:

\[ \int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a} + b - x} \, dx = \frac{b - a}{2}. \]

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