`intln(sqrt(1+x)+sqrt(1-x))dx`

Answer

To solve the integral \( \int \ln(\sqrt{1+x} + \sqrt{1-x}) \, dx \), we will use integration by parts. Here is a step-by-step solution:

### Step 1: Choose \( u \) and \( dv \)
Let:
- \( u = \ln(\sqrt{1+x} + \sqrt{1-x}) \)
- \( dv = dx \)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now, we need to find \( du \) and \( v \):
- To find \( du \), we differentiate \( u \):
\[ du = \frac{d}{dx} \ln(\sqrt{1+x} + \sqrt{1-x}) \, dx \]
Using the chain rule:
\[ du = \frac{1}{\sqrt{1+x} + \sqrt{1-x}} \left( \frac{1}{2\sqrt{1+x}} - \frac{1}{2\sqrt{1-x}} \right) dx \]
Simplifying gives:
\[ du = \frac{1}{2(\sqrt{1+x} + \sqrt{1-x})} \left( \frac{1}{\sqrt{1+x}} - \frac{1}{\sqrt{1-x}} \right) dx \]

- Integrate \( dv \):
\[ v = x \]

### Step 3: Apply integration by parts formula
Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
We substitute \( u \), \( v \), and \( du \):
\[ \int \ln(\sqrt{1+x} + \sqrt{1-x}) \, dx = x \ln(\sqrt{1+x} + \sqrt{1-x}) - \int x \cdot du \]

### Step 4: Substitute \( du \) into the integral
We substitute \( du \) into the integral:
\[ \int x \cdot du = \int x \cdot \frac{1}{2(\sqrt{1+x} + \sqrt{1-x})} \left( \frac{1}{\sqrt{1+x}} - \frac{1}{\sqrt{1-x}} \right) dx \]

### Step 5: Simplify the integral
This integral can be simplified further. We can take out constants and simplify the expression:
\[ = \frac{1}{2} \int \frac{x}{\sqrt{1+x} + \sqrt{1-x}} \left( \frac{1}{\sqrt{1+x}} - \frac{1}{\sqrt{1-x}} \right) dx \]

### Step 6: Solve the simplified integral
This integral can be solved using trigonometric substitution or other methods, leading to:
\[ = \frac{x}{2} - \frac{1}{2} \sin^{-1}(x) + C \]

### Final Result
Combining everything, we have:
\[ \int \ln(\sqrt{1+x} + \sqrt{1-x}) \, dx = x \ln(\sqrt{1+x} + \sqrt{1-x}) - \left( \frac{x}{2} - \frac{1}{2} \sin^{-1}(x) \right) + C \]