`((int_(-1/2)^(1/2)cos 2x *log((1+x)/(1-x))dx))/((int_(0)^(1/2)cos 2x *log((1+x)/(1-x))dx))` equals

Answer

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

\[ \frac{\int_{-\frac{1}{2}}^{\frac{1}{2}} \cos(2x) \log\left(\frac{1+x}{1-x}\right) \, dx}{\int_{0}^{\frac{1}{2}} \cos(2x) \log\left(\frac{1+x}{1-x}\right) \, dx} \]

### Step 1: Define the function
Let
\[ f(x) = \cos(2x) \log\left(\frac{1+x}{1-x}\right) \]

### Step 2: Check if the function is odd or even
To determine whether \( f(x) \) is odd or even, we need to compute \( f(-x) \):
\[ f(-x) = \cos(-2x) \log\left(\frac{1-x}{1+x}\right) \]

Using the property of cosine, \( \cos(-\theta) = \cos(\theta) \):
\[ f(-x) = \cos(2x) \log\left(\frac{1-x}{1+x}\right) \]

### Step 3: Simplify \( f(-x) \)
We can rewrite the logarithm:
\[ \log\left(\frac{1-x}{1+x}\right) = \log\left(\frac{1+x}{1-x}\right)^{-1} = -\log\left(\frac{1+x}{1-x}\right) \]

Thus, we have:
\[ f(-x) = \cos(2x) \left(-\log\left(\frac{1+x}{1-x}\right)\right) = -f(x) \]

### 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 from \(-\frac{1}{2}\) to \(\frac{1}{2}\)
For odd functions, the integral over symmetric limits around zero is zero:
\[ \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x) \, dx = 0 \]

### Step 6: Evaluate the denominator
The denominator is:
\[ \int_{0}^{\frac{1}{2}} f(x) \, dx \]
This integral is not zero, as it is the integral of an odd function over a non-symmetric interval.

### Step 7: Final result
Thus, the entire expression simplifies to:
\[ \frac{0}{\int_{0}^{\frac{1}{2}} f(x) \, dx} = 0 \]

Therefore, the final answer is:
\[ \boxed{0} \]