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. `2x+In|x+1|+c`
C. `3x+4In|x-1|+c`
D. `2x+3(n|x+1|+c`

Explanation

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

### Step-by-Step Solution:

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

2. **Expressing \( x \) in terms of \( t \)**:
From the substitution, we can rearrange it:
\[ t(x - 1) = x + 1 \implies tx - t = x + 1 \implies tx - x = t + 1 \implies x(t - 1) = t + 1 \implies x = \frac{t + 1}{t - 1} \]

3. **Finding \( f(t) \)**:
Substitute \( x = \frac{t + 1}{t - 1} \) into the equation \( f(t) = 2x + 1 \):
\[ f(t) = 2\left(\frac{t + 1}{t - 1}\right) + 1 = \frac{2(t + 1)}{t - 1} + 1 = \frac{2t + 2 + (t - 1)}{t - 1} = \frac{3t + 1}{t - 1} \]

4. **Integral of \( f(x) \)**:
Now we need to find the integral:
\[ \int f(x) \, dx = \int \frac{3x + 1}{x - 1} \, dx \]

5. **Simplifying the Integral**:
We can split the fraction:
\[ \frac{3x + 1}{x - 1} = 3 + \frac{4}{x - 1} \]
Thus, the integral becomes:
\[ \int \left(3 + \frac{4}{x - 1}\right) \, dx \]

6. **Calculating the Integral**:
Now we can integrate term by term:
\[ \int 3 \, dx + \int \frac{4}{x - 1} \, dx = 3x + 4 \ln |x - 1| + C \]

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