The value of the integral
`int_(-1)^(1)log(x+sqrt(x^(2)+1))dx` is

Options

A. 0
B. log2
C. `log1//2`
D. none of these

Explanation

To solve the integral

\[ I = \int_{-1}^{1} \log(x + \sqrt{x^2 + 1}) \, dx, \]

we will analyze the function inside the integral.

### Step 1: Define the function
Let

\[ f(x) = \log(x + \sqrt{x^2 + 1}). \]

### Step 2: Evaluate \( f(-x) \)
Next, we will find \( f(-x) \):

\[ f(-x) = \log(-x + \sqrt{(-x)^2 + 1}) = \log(-x + \sqrt{x^2 + 1}). \]

### Step 3: Add \( f(x) \) and \( f(-x) \)
Now, we add \( f(x) \) and \( f(-x) \):

\[ f(x) + f(-x) = \log(x + \sqrt{x^2 + 1}) + \log(-x + \sqrt{x^2 + 1}). \]

Using the property of logarithms that states \( \log(a) + \log(b) = \log(ab) \), we can combine the logs:

\[ f(x) + f(-x) = \log\left((x + \sqrt{x^2 + 1})(-x + \sqrt{x^2 + 1})\right). \]

### Step 4: Simplify the product
Now, we simplify the expression inside the logarithm:

\[ (x + \sqrt{x^2 + 1})(-x + \sqrt{x^2 + 1}) = \sqrt{x^2 + 1}^2 - x^2 = (x^2 + 1) - x^2 = 1. \]

Thus, we have:

\[ f(x) + f(-x) = \log(1) = 0. \]

### Step 5: Conclude that \( f(x) \) is odd
From the above result, we see that:

\[ f(x) + f(-x) = 0 \implies f(-x) = -f(x). \]

This means that \( f(x) \) is an odd function.

### Step 6: Evaluate the integral
Since \( f(x) \) is an odd function, we can use the property of definite integrals for odd functions:

\[ \int_{-a}^{a} f(x) \, dx = 0. \]

In our case, since we are integrating from \(-1\) to \(1\):

\[ I = \int_{-1}^{1} f(x) \, dx = 0. \]

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

\[ \boxed{0}. \]

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