`int x(ln(x+sqrt(1+x^2))/sqrt(1+x^2)dx`

Options

A. `[log (x+sqrt(1+x^2))]^2`
B. `frac {1}{4}[log (x+sqrt(1+x^2))]`
C. `log (x+sqrt(1+x^2))sqrt(1+x^2)-x+c`
D. None of these

Answer

To solve the integral \( I = \int x \frac{\ln(x + \sqrt{1+x^2})}{\sqrt{1+x^2}} \, dx \), we will use the substitution method and integration by parts. Here’s the step-by-step solution:

### Step 1: Substitution
Let \( x = \tan \theta \). Then, we have:
\[ dx = \sec^2 \theta \, d\theta \]
Also, we know:
\[ \sqrt{1+x^2} = \sqrt{1+\tan^2 \theta} = \sec \theta \]
Substituting these into the integral gives:
\[ I = \int \tan \theta \frac{\ln(\tan \theta + \sec \theta)}{\sec \theta} \sec^2 \theta \, d\theta \]
This simplifies to:
\[ I = \int \tan \theta \ln(\tan \theta + \sec \theta) \sec \theta \, d\theta \]

### Step 2: Integration by Parts
Let:
- \( u = \ln(\tan \theta + \sec \theta) \)
- \( dv = \tan \theta \sec \theta \, d\theta \)

Now, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{\tan \theta + \sec \theta} (\sec^2 \theta + \tan \theta \sec \theta) \, d\theta \]
- Integrate \( dv \):
\[ v = \sec \theta \]

Now applying integration by parts:
\[ I = u v - \int v \, du \]
This gives:
\[ I = \ln(\tan \theta + \sec \theta) \sec \theta - \int \sec \theta \cdot \frac{1}{\tan \theta + \sec \theta} (\sec^2 \theta + \tan \theta \sec \theta) \, d\theta \]

### Step 3: Simplifying the Integral
The integral can be simplified further, but for brevity, we will focus on the main result. The integration of \( \sec^2 \theta \) is straightforward:
\[ \int \sec^2 \theta \, d\theta = \tan \theta \]

### Step 4: Back Substitution
Now, we substitute back \( \theta \) in terms of \( x \):
\[ \sec \theta = \sqrt{1+x^2}, \quad \tan \theta = x \]
Thus, we have:
\[ I = \ln(x + \sqrt{1+x^2}) \sqrt{1+x^2} - \tan \theta + C \]
Substituting back gives:
\[ I = \ln(x + \sqrt{1+x^2}) \sqrt{1+x^2} - x + C \]

### Final Answer
The final expression for the integral is:
\[ I = \ln(x + \sqrt{1+x^2}) \sqrt{1+x^2} - x + C \]