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

Options

A. 0
B. `(1)/(2)`
C. `-(1)/(2)`
D. none of these

Explanation

To solve the integral

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

we will determine whether the integrand is an odd or even function.

### Step 1: Identify the function
Let

\[ f(x) = \cos x \log\left(\frac{1+x}{1-x}\right). \]

### Step 2: Check if the function is odd or even
To check if \( f(x) \) is odd, we compute \( f(-x) \):

\[ f(-x) = \cos(-x) \log\left(\frac{1-x}{1+x}\right). \]

Using the property \( \cos(-x) = \cos x \), we have:

\[ f(-x) = \cos x \log\left(\frac{1-x}{1+x}\right). \]

### Step 3: Simplify \( f(-x) \)
We can rewrite the logarithm:

\[ \log\left(\frac{1-x}{1+x}\right) = \log(1-x) - \log(1+x). \]

Thus,

\[ f(-x) = \cos x \left(\log(1-x) - \log(1+x)\right). \]

Using the property of logarithms, we can express this as:

\[ f(-x) = \cos x \left(-\log\left(\frac{1+x}{1-x}\right)\right) = -\cos x \log\left(\frac{1+x}{1-x}\right) = -f(x). \]

### Step 4: Conclusion about the function
Since \( f(-x) = -f(x) \), we conclude that \( f(x) \) is an odd function.

### Step 5: Evaluate the integral
Using the property of integrals, if \( f(x) \) is odd, then:

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

In our case, since the limits are from \(-\frac{1}{2}\) to \(\frac{1}{2}\):

\[ I = \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \, dx = 0. \]

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

\[ \boxed{0}. \]