Evaluate the following integrals:
`int_(-1/2)^(1/2) cos x log ((1+x)/(1-x))dx`

Answer

To evaluate the integral

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

we will check if the integrand is an odd or even function.

### Step 1: Define the function
Let

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

### Step 2: Check if \( f(x) \) is odd or even
To determine if \( f(x) \) is odd, we need to compute \( f(-x) \):

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

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

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

### Step 3: Simplify \( f(-x) \)
Now, we can rewrite the logarithm:

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

Thus,

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

### 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
By the property of integrals of odd functions over symmetric intervals, we have:

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

Therefore,

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

### Final Answer
The value of the integral is

\[ \boxed{0}. \]