if ` f ((x- 4 ) /(x + 2 )) = 2 x + 1 , (x in R - { 1, - 2 }) ` , then ` int f(x) dx ` is equal to : (where C is constant of integration)

Options

A. `12 log _e | 1 - x | - 3 x + c `
B. ` -12 log _ e | 1 - x | - 3x + c `
C. ` -12 log _e | 1 - x | + 3x +c `
D. ` 12 log _e| 1 - x | + 3x + c `

Answer

To solve the problem, we need to find the integral of the function \( f(x) \) given that \( f\left(\frac{x-4}{x+2}\right) = 2x + 1 \).

### Step-by-Step Solution:

1. **Substitution**: Let \( y = \frac{x-4}{x+2} \). This will help us express \( x \) in terms of \( y \).

\[ y(x + 2) = x - 4 \implies yx + 2y = x - 4 \implies yx - x = -4 - 2y \implies x(y - 1) = -4 - 2y \]

Thus, we can express \( x \) as:

\[ x = \frac{-4 - 2y}{y - 1} \]

2. **Finding \( f(y) \)**: Substitute \( x \) back into the equation \( f(y) = 2x + 1 \):

\[ f(y) = 2\left(\frac{-4 - 2y}{y - 1}\right) + 1 \]

Simplifying this:

\[ f(y) = \frac{-8 - 4y}{y - 1} + 1 = \frac{-8 - 4y + (y - 1)}{y - 1} = \frac{-9 - 3y}{y - 1} \]

3. **Final Form of \( f(x) \)**: Now we can express \( f(x) \) as:

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

4. **Integrating \( f(x) \)**: Now we need to integrate \( f(x) \):

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

We can separate this into two integrals:

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

5. **Calculating the Integrals**:
- The first integral:
\[ \int \frac{3}{x - 1} \, dx = 3 \ln |x - 1| \]
- The second integral:
\[ \int 1 \, dx = x \]

Thus, we have:

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

6. **Final Result**:
\[ \int f(x) \, dx = -9 \ln |x - 1| - 3x + C \]

### Conclusion:
The integral \( \int f(x) \, dx \) is equal to \( -9 \ln |x - 1| - 3x + C \).