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

Options

A. 0
B. log 2
C. log (1//2)
D. None of these

Answer

To solve the integral \( I = \int_{-1}^{1} \log(x + \sqrt{x^2 + 1}) \, dx \), we will use the property of definite integrals and the symmetry of the function involved.

### Step 1: Define the function
Let
\[ f(x) = \log(x + \sqrt{x^2 + 1}). \]

### Step 2: Find \( f(-x) \)
Now, we will compute \( f(-x) \):
\[ f(-x) = \log(-x + \sqrt{(-x)^2 + 1}) = \log(-x + \sqrt{x^2 + 1}). \]

### Step 3: Simplify \( f(-x) \)
We can rewrite \( f(-x) \) using properties of logarithms:
\[ f(-x) = \log\left(\sqrt{x^2 + 1} - x\right). \]

### Step 4: Relate \( f(-x) \) to \( f(x) \)
Next, we can express \( f(-x) \) in terms of \( f(x) \):
\[ f(-x) = \log\left(\frac{1}{x + \sqrt{x^2 + 1}}\right) = -\log(x + \sqrt{x^2 + 1}) = -f(x). \]
This shows that:
\[ f(-x) = -f(x). \]

### Step 5: Apply the property of definite integrals
Since \( f(-x) = -f(x) \), we can use the property of definite integrals:
\[ \int_{-a}^{a} f(x) \, dx = 0 \text{ if } f(-x) = -f(x). \]
Thus, we have:
\[ \int_{-1}^{1} f(x) \, dx = \int_{-1}^{1} \log(x + \sqrt{x^2 + 1}) \, dx = 0. \]

### Conclusion
Therefore, the value of the integral is:
\[ \int_{-1}^{1} \log(x + \sqrt{x^2 + 1}) \, dx = 0. \]