The value
`int^(2)_(-2) {p" In"((1+x)/(1-x))+q" In "((1-x)/(1+x))-2+r}dx` depends on the value of

Options

A. p
B. q
C. r
D. p and q

Explanation

To solve the integral

\[ I = \int_{-2}^{2} \left( p \ln \left( \frac{1+x}{1-x} \right) - 2q \ln \left( \frac{1-x}{1+x} \right) + r \right) dx, \]

we will break it down step by step.

### Step 1: Rewrite the Integral
We start by rewriting the integral:

\[ I = \int_{-2}^{2} \left( p \ln \left( \frac{1+x}{1-x} \right) + r - 2q \ln \left( \frac{1-x}{1+x} \right) \right) dx. \]

### Step 2: Use Logarithmic Properties
Using the property of logarithms, we can rewrite the second logarithmic term:

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

Thus, we can express the integral as:

\[ I = \int_{-2}^{2} \left( p \ln \left( \frac{1+x}{1-x} \right) + r + 2q \ln \left( \frac{1+x}{1-x} \right) \right) dx. \]

### Step 3: Combine the Logarithmic Terms
Now we combine the logarithmic terms:

\[ I = \int_{-2}^{2} \left( (p + 2q) \ln \left( \frac{1+x}{1-x} \right) + r \right) dx. \]

### Step 4: Separate the Integral
We can separate the integral into two parts:

\[ I = (p + 2q) \int_{-2}^{2} \ln \left( \frac{1+x}{1-x} \right) dx + \int_{-2}^{2} r \, dx. \]

### Step 5: Evaluate the Integral of r
The integral of \( r \) is straightforward:

\[ \int_{-2}^{2} r \, dx = r \cdot (2 - (-2)) = 4r. \]

### Step 6: Determine the Nature of the Logarithmic Integral
Next, we need to evaluate

\[ \int_{-2}^{2} \ln \left( \frac{1+x}{1-x} \right) dx. \]

To analyze this integral, we check if the function is odd or even.

Let

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

Calculating \( f(-x) \):

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

Since \( f(-x) = -f(x) \), \( f(x) \) is an odd function. Therefore, the integral over a symmetric interval around zero is zero:

\[ \int_{-2}^{2} \ln \left( \frac{1+x}{1-x} \right) dx = 0. \]

### Step 7: Final Expression for I
Substituting this back into our expression for \( I \):

\[ I = (p + 2q) \cdot 0 + 4r = 4r. \]

### Conclusion
Thus, the value of the integral depends solely on \( r \).

The final answer is:

\[ \text{The value of the integral depends on } r. \]