The value of `int_(1//n)^((a n-1)//n) (sqrtx)/(sqrt(a-x) + sqrtx)dx` is equal to

Options

A. `(a)/(2)`
B. `(1)/(2n) (n a+2)`
C. `(n a-2)/(2n)`
D. None of these

Answer

To solve the integral

\[ I = \int_{\frac{1}{n}}^{\frac{a n - 1}{n}} \frac{\sqrt{x}}{\sqrt{a - x} + \sqrt{x}} \, dx, \]

we will use a property of definite integrals. This property states that:

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

### Step 1: Define the Integral
Let's define the integral as:

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

### Step 2: Change of Variable
Using the property mentioned, we will change the variable in the integral. The new variable will be \( a + b - x \), where \( a = \frac{1}{n} \) and \( b = \frac{a n - 1}{n} \).

Calculating \( a + b \):

\[ a + b = \frac{1}{n} + \frac{a n - 1}{n} = \frac{a n}{n} = a. \]

So we have:

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

### Step 3: Combine Integrals
Now we will add this new integral to the original integral \( I \):

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

### Step 4: Simplify the Expression
The expression inside the integral simplifies to:

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

Thus, we have:

\[ 2I = \int_{\frac{1}{n}}^{\frac{a n - 1}{n}} 1 \, dx. \]

### Step 5: Evaluate the Integral
Now we can evaluate the integral:

\[ 2I = \left[ x \right]_{\frac{1}{n}}^{\frac{a n - 1}{n}} = \frac{a n - 1}{n} - \frac{1}{n} = \frac{a n - 2}{n}. \]

### Step 6: Solve for \( I \)
Now, we can solve for \( I \):

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

### Final Answer
Thus, the value of the integral is:

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