Using properties of definite integration evaluate : `int_(a)^(b) (root(n)x)/(root(n)x+root(n)(a+b-x))dx`

Answer

To evaluate the integral

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

we can use the property of definite integrals that states:

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

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

Let’s denote the integral as \(I\):

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

Now, we will change the variable in the integral using the property mentioned above:

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

### Step 2: Simplify the new integral

The new integral can be simplified as follows:

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

### Step 3: Combine the two integrals

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

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

Adding these two equations:

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

### Step 4: Simplify the sum

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

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

This simplifies to:

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

### Step 5: Evaluate the integral

Now we can evaluate the integral:

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

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

Thus, we have:

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

### Final Answer

The value of the integral is:

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