The value of `intlog (sqrt(1- x) + sqrt(1+ x)) dx, ` is equal to

Options

A. `x log (sqrt(1-x)+sqrt(1+x))+1/2 x - 1/2 sin^(-1)(x)+C`
B. `x log (sqrt(1-x)+sqrt(1+x))+1/2 x + 1/2 sin^(-1)(x)+C`
C. `x log (sqrt(1-x)+sqrt(1+x))- 1/2 x + 1/2 sin^(-1)(x)+C`
D. None of the above

Answer

To solve the integral \( I = \int \log(\sqrt{1 - x} + \sqrt{1 + x}) \, dx \), we will follow a systematic approach using integration by parts.

### Step-by-Step Solution:

1. **Set Up the Integral**:
\[ I = \int \log(\sqrt{1 - x} + \sqrt{1 + x}) \, dx \]

2. **Use Integration by Parts**:
We choose:
- \( u = \log(\sqrt{1 - x} + \sqrt{1 + x}) \)
- \( dv = dx \)

Then, we differentiate \( u \) and integrate \( dv \):
- \( du = \frac{1}{\sqrt{1 - x} + \sqrt{1 + x}} \left( \frac{-1}{2\sqrt{1 - x}} + \frac{1}{2\sqrt{1 + x}} \right) \, dx \)
- \( v = x \)

3. **Apply Integration by Parts Formula**:
The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du \]
Thus, we have:
\[ I = x \log(\sqrt{1 - x} + \sqrt{1 + x}) - \int x \cdot du \]

4. **Calculate \( \int x \, du \)**:
Substitute \( du \):
\[ I = x \log(\sqrt{1 - x} + \sqrt{1 + x}) - \int x \cdot \left( \frac{1}{\sqrt{1 - x} + \sqrt{1 + x}} \left( \frac{-1}{2\sqrt{1 - x}} + \frac{1}{2\sqrt{1 + x}} \right) \right) \, dx \]

5. **Simplify the Integral**:
This integral can be simplified further. We can express it as:
\[ I = x \log(\sqrt{1 - x} + \sqrt{1 + x}) + \frac{1}{2} \int \frac{x}{\sqrt{1 - x} + \sqrt{1 + x}} \left( \frac{1}{\sqrt{1 - x}} + \frac{1}{\sqrt{1 + x}} \right) \, dx \]

6. **Evaluate the Remaining Integral**:
The remaining integral can be evaluated using substitution or recognizing it as a standard form. After simplification, we find:
\[ \int \frac{1}{\sqrt{1 - x^2}} \, dx = \sin^{-1}(x) \]

7. **Combine Results**:
Therefore, we can combine all parts:
\[ I = x \log(\sqrt{1 - x} + \sqrt{1 + x}) + \sin^{-1}(x) - x + C \]

### Final Answer:
\[ I = x \log(\sqrt{1 - x} + \sqrt{1 + x}) + \sin^{-1}(x) - x + C \]