If `intsqrt(((1-sqrtx))/(1+sqrtx))dx=Asqrt(1-x)+B sin^(-1)sqrtx+Csqrt(x-x^(2))+D`, where `A+B+C=`

Options

A. `-1`
B. `-2`
C. 3
D. `-4`

Answer

To solve the integral \( \int \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \, dx \) and express it in the form \( A\sqrt{1 - x} + B \sin^{-1}(\sqrt{x}) + C\sqrt{x - x^2} + D \), we will follow these steps:

### Step 1: Substitution
Let \( \sqrt{x} = \sin(t) \). Then, \( x = \sin^2(t) \) and \( dx = 2\sin(t)\cos(t) \, dt \).

### Step 2: Rewrite the Integral
Substituting into the integral gives:
\[ \int \sqrt{\frac{1 - \sin(t)}{1 + \sin(t)}} \cdot 2\sin(t)\cos(t) \, dt \]

### Step 3: Simplify the Expression
We can simplify \( \sqrt{\frac{1 - \sin(t)}{1 + \sin(t)}} \):
\[ \sqrt{\frac{1 - \sin(t)}{1 + \sin(t)}} = \frac{\sqrt{(1 - \sin(t))(1 - \sin(t))}}{\sqrt{(1 + \sin(t))(1 - \sin(t))}} = \frac{1 - \sin(t)}{\sqrt{1 - \sin^2(t)}} = \frac{1 - \sin(t)}{\cos(t)} \]
Thus, the integral becomes:
\[ \int 2\sin(t)\cos(t) \cdot \frac{1 - \sin(t)}{\cos(t)} \, dt = \int 2\sin(t)(1 - \sin(t)) \, dt \]

### Step 4: Expand the Integral
Expanding the integrand:
\[ \int 2\sin(t) - 2\sin^2(t) \, dt \]

### Step 5: Integrate Each Term
The integral can be split:
\[ \int 2\sin(t) \, dt - \int 2\sin^2(t) \, dt \]
The first integral is:
\[ -2\cos(t) \]
For the second integral, we use the identity \( \sin^2(t) = \frac{1 - \cos(2t)}{2} \):
\[ \int 2\sin^2(t) \, dt = \int (1 - \cos(2t)) \, dt = t - \frac{1}{2}\sin(2t) \]

### Step 6: Combine Results
Combining these results gives:
\[ -2\cos(t) - \left(t - \frac{1}{2}\sin(2t)\right) + C \]
Substituting back \( t = \sin^{-1}(\sqrt{x}) \) and \( \cos(t) = \sqrt{1 - x} \):
\[ -2\sqrt{1 - x} - \sin^{-1}(\sqrt{x}) + \sin(t)\cos(t) + C \]

### Step 7: Final Form
Thus, we can express the result as:
\[ -2\sqrt{1 - x} - \sin^{-1}(\sqrt{x}) + \sqrt{x(1 - x)} + C \]
Comparing with \( A\sqrt{1 - x} + B \sin^{-1}(\sqrt{x}) + C\sqrt{x - x^2} + D \), we find:
- \( A = -2 \)
- \( B = -1 \)
- \( C = 1 \)

### Step 8: Calculate \( A + B + C \)
Now, we calculate:
\[ A + B + C = -2 - 1 + 1 = -2 \]

### Final Answer
Thus, the value of \( A + B + C \) is \( \boxed{-2} \).