The value of `int(log_(e)(x+sqrt(x^2)+1))/(sqrt(x^(2)+1))dx` is

Options

A. `2 log_(e)(x+sqrt(x^(2)+1))+C`
B. `{log_(e)(x+sqrt(x^(2)+1))}^(2)+C`
C. `log(x+sqrt(x^(2)+1))+C`
D. none of these

Explanation

To solve the integral

\[ I = \int \frac{\ln(x + \sqrt{x^2 + 1})}{\sqrt{x^2 + 1}} \, dx, \]

we will use a substitution method. Let's follow the steps:

### Step 1: Substitution
Let

\[ t = x + \sqrt{x^2 + 1}. \]

### Step 2: Differentiate the substitution
Now, we differentiate \(t\) with respect to \(x\):

\[ dt = \left(1 + \frac{x}{\sqrt{x^2 + 1}}\right) dx. \]

### Step 3: Simplify the derivative
We can rewrite the derivative:

\[ dt = \frac{\sqrt{x^2 + 1} + x}{\sqrt{x^2 + 1}} \, dx = \frac{t}{\sqrt{x^2 + 1}} \, dx. \]

### Step 4: Solve for \(dx\)
From this, we can express \(dx\):

\[ dx = \frac{\sqrt{x^2 + 1}}{t} \, dt. \]

### Step 5: Substitute back into the integral
Now, substituting \(t\) and \(dx\) into the integral:

\[ I = \int \frac{\ln(t)}{\sqrt{x^2 + 1}} \cdot \frac{\sqrt{x^2 + 1}}{t} \, dt = \int \frac{\ln(t)}{t} \, dt. \]

### Step 6: Integrate
The integral of \(\frac{\ln(t)}{t}\) is:

\[ \int \frac{\ln(t)}{t} \, dt = \frac{(\ln(t))^2}{2} + C. \]

### Step 7: Back substitute for \(t\)
Now, we substitute back \(t = x + \sqrt{x^2 + 1}\):

\[ I = \frac{(\ln(x + \sqrt{x^2 + 1}))^2}{2} + C. \]

### Final Answer
Thus, the value of the integral is:

\[ I = \frac{1}{2} \ln^2(x + \sqrt{x^2 + 1}) + C. \]

---