Let ` f ((x + 1) /(x - 1)) = 2x + 1 `, then integral `int f(x)\ dx` is ` (x ne 1)`

Options

A. ` x ^ 2 + x + c `
B. ` 2 x + ln | x + 1 | + c `
C. ` 3x + 4ln | x - 1 | + c `
D. `2 x + 3 ln | x + 1 | + c `

Explanation

To solve the problem, we need to find the integral of the function \( f(x) \) given that \( f\left(\frac{x + 1}{x - 1}\right) = 2x + 1 \).

### Step-by-Step Solution:

1. **Substitution**: Let \( t = \frac{x + 1}{x - 1} \). We will express \( x \) in terms of \( t \).

\[ t(x - 1) = x + 1 \implies tx - t = x + 1 \implies tx - x = t + 1 \implies x(t - 1) = t + 1 \implies x = \frac{t + 1}{t - 1} \]

2. **Finding \( 2x + 1 \)**: Now, substitute \( x = \frac{t + 1}{t - 1} \) into \( 2x + 1 \).

\[ 2x + 1 = 2\left(\frac{t + 1}{t - 1}\right) + 1 = \frac{2(t + 1)}{t - 1} + 1 = \frac{2t + 2 + t - 1}{t - 1} = \frac{3t + 1}{t - 1} \]

3. **Identifying \( f(t) \)**: Since \( f\left(\frac{x + 1}{x - 1}\right) = 2x + 1 \), we have:

\[ f(t) = 3 + \frac{4}{t - 1} \]

4. **Finding \( f(x) \)**: Replace \( t \) with \( x \) to find \( f(x) \).

\[ f(x) = 3 + \frac{4}{x - 1} \]

5. **Integrating \( f(x) \)**: Now we can integrate \( f(x) \).

\[ \int f(x) \, dx = \int \left(3 + \frac{4}{x - 1}\right) \, dx \]

- The integral of \( 3 \) is \( 3x \).
- The integral of \( \frac{4}{x - 1} \) is \( 4 \ln |x - 1| \).

Therefore,

\[ \int f(x) \, dx = 3x + 4 \ln |x - 1| + C \]

### Final Result:
The integral \( \int f(x) \, dx \) is given by:

\[ \int f(x) \, dx = 3x + 4 \ln |x - 1| + C \quad (x \neq 1) \]