Let `I =int _(0)^(1 ) sqrt((1+sqrtx)/(1-sqrtx))` dx then correct statement (s) is/are:

Options

A. `I=2`
B. `I =pi`
C. `I =2+(pi)/(2)`
D. `I =(4-pi)/(2)`

Answer

To solve the integral \( I = \int_0^1 \sqrt{\frac{1 + \sqrt{x}}{1 - \sqrt{x}}} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_0^1 \sqrt{\frac{1 + \sqrt{x}}{1 - \sqrt{x}}} \, dx \]

### Step 2: Rationalize the Expression
To simplify the integrand, we can rationalize it:
\[ \sqrt{\frac{1 + \sqrt{x}}{1 - \sqrt{x}}} = \frac{\sqrt{1 + \sqrt{x}}}{\sqrt{1 - \sqrt{x}}} \]
Thus, we can rewrite the integral as:
\[ I = \int_0^1 \frac{\sqrt{1 + \sqrt{x}}}{\sqrt{1 - \sqrt{x}}} \, dx \]

### Step 3: Substitute \( u = \sqrt{x} \)
Let \( u = \sqrt{x} \), then \( x = u^2 \) and \( dx = 2u \, du \). The limits change from \( x = 0 \) to \( x = 1 \) which corresponds to \( u = 0 \) to \( u = 1 \). The integral becomes:
\[ I = \int_0^1 \frac{\sqrt{1 + u}}{\sqrt{1 - u}} \cdot 2u \, du = 2 \int_0^1 \frac{u \sqrt{1 + u}}{\sqrt{1 - u}} \, du \]

### Step 4: Split the Integral
We can split the integral into two parts:
\[ I = 2 \left( \int_0^1 \frac{u}{\sqrt{1 - u}} \, du + \int_0^1 \frac{u^{3/2}}{\sqrt{1 - u}} \, du \right) \]

### Step 5: Evaluate the First Integral
The first integral can be evaluated using the substitution \( v = 1 - u \):
\[ \int_0^1 \frac{u}{\sqrt{1 - u}} \, du = \int_0^1 \frac{1 - v}{\sqrt{v}} \, dv = \int_0^1 \left( \frac{1}{\sqrt{v}} - \sqrt{v} \right) \, dv \]
Calculating this gives:
\[ = \left[ 2\sqrt{v} - \frac{2}{3}v^{3/2} \right]_0^1 = 2 - \frac{2}{3} = \frac{4}{3} \]

### Step 6: Evaluate the Second Integral
For the second integral:
\[ \int_0^1 \frac{u^{3/2}}{\sqrt{1 - u}} \, du \]
Using the same substitution \( v = 1 - u \):
\[ = \int_0^1 (1 - v)^{3/2} v^{-1/2} \, dv \]
This can be computed using the beta function or directly:
\[ = \frac{8}{15} \]

### Step 7: Combine the Results
Now we can combine the results:
\[ I = 2 \left( \frac{4}{3} + \frac{8}{15} \right) = 2 \left( \frac{20}{15} + \frac{8}{15} \right) = 2 \cdot \frac{28}{15} = \frac{56}{15} \]

### Final Result
Thus, the value of the integral \( I \) is:
\[ I = \frac{56}{15} \]