The value of
`int_(-1)^(1) (log(x+sqrt(1+x^(2))))/(x+log(x+sqrt(1+x^(2))))f(x) dx-int_(-1)^(1) (log(x +sqrt(1+x^(2))))/(x+log(x+sqrt(1+x^(2))))f(-x)dx`,

Options

A. 0
B. `2underset(0)overset(1)int (log(x+sqrt(1+x^(2))))/(x+log(x+sqrt(1+x^(2)))){f(x) -f(-x)}dx`
C. 2f(x)
D. none of these

Explanation

To solve the given problem, we need to evaluate the expression:

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

### Step 1: Define the function \( g(x) \)

Let

\[ g(x) = \frac{\log(x + \sqrt{1 + x^2})}{x + \log(x + \sqrt{1 + x^2})} \]

### Step 2: Rewrite the integrals

We can rewrite the expression \( I \) as:

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

### Step 3: Combine the integrals

Combine the two integrals:

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

### Step 4: Analyze \( f(x) - f(-x) \)

Define

\[ h(x) = f(x) - f(-x) \]

### Step 5: Check if \( h(x) \) is odd

To check if \( h(x) \) is odd, we compute \( h(-x) \):

\[ h(-x) = f(-x) - f(x) = - (f(x) - f(-x)) = -h(x) \]

Thus, \( h(x) \) is an odd function.

### Step 6: Check if \( g(x) \) is even or odd

Now, we need to check if \( g(x) \) is even or odd. We compute \( g(-x) \):

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

Notice that:

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

Thus,

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

### Step 7: Simplify \( g(x) - g(-x) \)

We find that:

\[ g(-x) = -g(x) \]

This shows that \( g(x) \) is an odd function.

### Step 8: Combine the properties of \( g(x) \) and \( h(x) \)

Since \( g(x) \) is odd and \( h(x) \) is odd, the product \( g(x) h(x) \) is even.

### Step 9: Evaluate the integral

Now we can evaluate the integral:

\[ I = \int_{-1}^{1} g(x) h(x) \, dx \]

Since \( g(x) h(x) \) is an odd function, the integral over a symmetric interval around zero will be zero:

\[ I = 0 \]

### Final Answer

Thus, the value of the given expression is:

\[ \boxed{0} \]