If `int(1)/((x^(2)-1))ln((x-1)/(x+1))dx=6A[ln ((x-1)/(x+1))]^(2)+C`, then find 24 A.

Answer

To solve the integral

\[ \int \frac{1}{x^2 - 1} \ln\left(\frac{x-1}{x+1}\right) \, dx = 6A \left[\ln\left(\frac{x-1}{x+1}\right)\right]^2 + C, \]

we will follow these steps:

### Step 1: Substitution
Let
\[ t = \ln\left(\frac{x-1}{x+1}\right). \]
Then, we need to find \( dt \).

### Step 2: Differentiate \( t \)
Using the chain rule, we differentiate \( t \):
\[ \frac{d}{dx}\left(\ln\left(\frac{x-1}{x+1}\right)\right) = \frac{1}{\frac{x-1}{x+1}} \cdot \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} = \frac{2}{(x-1)(x+1)} = \frac{2}{x^2 - 1}. \]
Thus, we have:
\[ dt = \frac{2}{x^2 - 1} \, dx \implies dx = \frac{(x^2 - 1)}{2} \, dt. \]

### Step 3: Substitute in the Integral
Substituting \( t \) and \( dx \) into the integral:
\[ \int \frac{1}{x^2 - 1} \ln\left(\frac{x-1}{x+1}\right) \, dx = \int \frac{1}{x^2 - 1} \cdot t \cdot \frac{(x^2 - 1)}{2} \, dt. \]
This simplifies to:
\[ \int \frac{t}{2} \, dt = \frac{1}{2} \int t \, dt. \]

### Step 4: Integrate
Now we integrate:
\[ \frac{1}{2} \int t \, dt = \frac{1}{2} \cdot \frac{t^2}{2} + C = \frac{t^2}{4} + C. \]

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

### Step 6: Compare with Given Expression
We have:
\[ \frac{1}{4} \left[\ln\left(\frac{x-1}{x+1}\right)\right]^2 + C = 6A \left[\ln\left(\frac{x-1}{x+1}\right)\right]^2 + C. \]
This implies:
\[ 6A = \frac{1}{4} \implies A = \frac{1}{24}. \]

### Step 7: Find \( 24A \)
Now, we calculate:
\[ 24A = 24 \cdot \frac{1}{24} = 1. \]

Thus, the final answer is:
\[ \boxed{1}. \]