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:

**Step 1: Substitute the argument of \( f \)**
Let \( X = \frac{x - 4}{x + 2} \). We need to express \( x \) in terms of \( X \).

**Hint:** To find \( x \) in terms of \( X \), rearrange the equation \( X(x + 2) = x - 4 \).

**Step 2: Rearranging the equation**
From \( X(x + 2) = x - 4 \), we can rewrite it as:
\[ Xx + 2X = x - 4 \]
Rearranging gives:
\[ Xx - x = -4 - 2X \]
Factoring out \( x \):
\[ x(X - 1) = -4 - 2X \]
Thus,
\[ x = \frac{-4 - 2X}{X - 1} \]

**Hint:** Make sure to simplify the expression correctly.

**Step 3: Substitute \( x \) back into the function**
Now substitute \( x \) back into the function \( f \):
\[ f(X) = 2\left(\frac{-4 - 2X}{X - 1}\right) + 1 \]
This simplifies to:
\[ f(X) = \frac{-8 - 4X}{X - 1} + 1 \]

**Hint:** Combine the terms carefully, ensuring you have a common denominator.

**Step 4: Simplifying \( f(X) \)**
Combining the terms gives:
\[ f(X) = \frac{-8 - 4X + (X - 1)}{X - 1} = \frac{-8 - 4X + X - 1}{X - 1} = \frac{-3X - 9}{X - 1} \]

**Hint:** Factor out common terms if possible.

**Step 5: Integrating \( f(X) \)**
Now we need to integrate \( f(X) \):
\[ \int f(X) \, dX = \int \left(-3 + \frac{12}{1 - X}\right) dX \]

**Hint:** Break the integral into two parts for easier integration.

**Step 6: Performing the integration**
Integrating gives:
\[ -3X - 12 \ln |1 - X| + C \]

**Hint:** Remember to include the constant of integration \( C \).

### Final Answer:
Thus, the integral \( \int f(x) \, dx \) is equal to:
\[ -3X - 12 \ln |1 - X| + C \]

Where \( X = \frac{x - 4}{x + 2} \).