`int("In"((x-1)/(x+1)))/(x^(2)-1)dx` is equal to

Options

A. `(1)/(2)("In"((x-1)/(x+1)))^(2)+C`
B. `(1)/(2)("In"((x+1)/(x-1)))^(2)+C`
C. `(1)/(4)("In"((x-1)/(x+1)))^(2)+C`
D. `(1)/(4)("In"((x+1)/(x-1)))`

Explanation

To solve the integral \( I = \int \frac{\ln\left(\frac{x-1}{x+1}\right)}{x^2 - 1} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\ln\left(\frac{x-1}{x+1}\right)}{x^2 - 1} \, dx \]
Notice that \( x^2 - 1 \) can be factored as \( (x-1)(x+1) \).

### Step 2: Substitution
Let:
\[ t = \ln\left(\frac{x-1}{x+1}\right) \]
To differentiate \( t \), we use the quotient rule:
\[ \frac{d}{dx}\left(\ln\left(\frac{x-1}{x+1}\right)\right) = \frac{(x+1)(1) - (x-1)(1)}{(x-1)(x+1)} = \frac{(x+1) - (x-1)}{(x-1)(x+1)} = \frac{2}{(x-1)(x+1)} = \frac{2}{x^2 - 1} \]
Thus, we have:
\[ dt = \frac{2}{x^2 - 1} \, dx \quad \Rightarrow \quad dx = \frac{x^2 - 1}{2} \, dt \]

### Step 3: Substitute in the Integral
Now we can substitute back into the integral:
\[ I = \int t \cdot \frac{x^2 - 1}{2} \cdot \frac{1}{x^2 - 1} \, dt = \frac{1}{2} \int t \, dt \]

### Step 4: Integrate
Integrating \( \frac{1}{2} \int t \, dt \):
\[ I = \frac{1}{2} \cdot \frac{t^2}{2} + C = \frac{t^2}{4} + C \]

### Step 5: Substitute Back for \( t \)
Now substitute back \( t = \ln\left(\frac{x-1}{x+1}\right) \):
\[ I = \frac{1}{4} \left(\ln\left(\frac{x-1}{x+1}\right)\right)^2 + C \]

### Final Answer
Thus, the final answer is:
\[ I = \frac{1}{4} \left(\ln\left(\frac{x-1}{x+1}\right)\right)^2 + C \]