`int _(-1//2)^(1//2) cos x log ((1+x)/(1-x)) dx = k log 2 , ` then k equals

Options

A. 0
B. `-1`
C. `-2`
D. `1/2`

Explanation

To solve the integral

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

we will analyze the integrand and determine if it 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 \( f(x) \) is odd or even
To determine if \( f(x) \) is odd or even, we need to compute \( f(-x) \):

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

Using the properties of cosine and logarithm, we have:

\[ \cos(-x) = \cos x, \]
and

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

Thus,

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

### Step 3: Conclude that \( f(x) \) is odd
Since \( f(-x) = -f(x) \), we conclude that \( f(x) \) is an odd function.

### Step 4: Evaluate the integral of an odd function over symmetric limits
The integral of an odd function over a symmetric interval around zero is zero. Therefore,

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

### Step 5: Relate to the given equation
Given that

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

and we found that the integral equals zero, we have:

\[ 0 = k \log 2. \]

### Step 6: Solve for \( k \)
Since \( \log 2 \neq 0 \), we can conclude that:

\[ k = 0. \]

Thus, the value of \( k \) is

\[ \boxed{0}. \]