If `f((3x-4)/(3x+4))=x+2`, then `int` f(x)dx is equal to

Options

A. `e^(x+2)log_(e)|(3x-4)/(3x+4)|`
B. `-(8)/(3)log_(e)|1-x|+(2)/(3)x+C`
C. `(8)/(3)log_(e)|x-1|+(x)/(3)+C`
D. none of these

Explanation

To solve the problem, we need to find the indefinite integral of the function \( f(x) \) given that \( f\left(\frac{3x-4}{3x+4}\right) = x + 2 \).

### Step-by-Step Solution:

1. **Substitution**: Let \( y = \frac{3x - 4}{3x + 4} \). We need to express \( x \) in terms of \( y \).
\[ y(3x + 4) = 3x - 4 \implies 3xy + 4y = 3x - 4 \]
Rearranging gives:
\[ 3xy - 3x = -4 - 4y \implies 3x(y - 1) = -4(1 + y) \implies x = \frac{-4(1 + y)}{3(y - 1)} \]

2. **Finding \( f(y) \)**: From the original equation, we have:
\[ f(y) = x + 2 = \frac{-4(1 + y)}{3(y - 1)} + 2 \]
To simplify this, we first express \( 2 \) with a common denominator:
\[ 2 = \frac{6(y - 1)}{3(y - 1)} = \frac{6y - 6}{3(y - 1)} \]
Now substituting back:
\[ f(y) = \frac{-4(1 + y) + 6y - 6}{3(y - 1)} = \frac{-4 - 4y + 6y - 6}{3(y - 1)} = \frac{2y - 10}{3(y - 1)} \]

3. **Simplifying \( f(y) \)**:
\[ f(y) = \frac{2(y - 5)}{3(y - 1)} \]

4. **Finding the integral**: Now we need to integrate \( f(x) \):
\[ \int f(x) \, dx = \int \frac{2(x - 5)}{3(x - 1)} \, dx \]
This can be split into two parts:
\[ \int \frac{2x}{3(x - 1)} \, dx - \int \frac{10}{3(x - 1)} \, dx \]

5. **Integrating the first term**: For the first term, we can use substitution:
\[ u = x - 1 \implies du = dx \implies x = u + 1 \]
Thus,
\[ \int \frac{2(u + 1)}{3u} \, du = \int \left(\frac{2}{3} + \frac{2}{3u}\right) \, du = \frac{2}{3}u + \frac{2}{3} \ln |u| + C \]

6. **Integrating the second term**:
\[ -\int \frac{10}{3(x - 1)} \, dx = -\frac{10}{3} \ln |x - 1| + C \]

7. **Combining the results**:
\[ \int f(x) \, dx = \frac{2}{3}(x - 1) + \frac{2}{3} \ln |x - 1| - \frac{10}{3} \ln |x - 1| + C \]
Simplifying gives:
\[ \int f(x) \, dx = \frac{2}{3}(x - 1) - \frac{8}{3} \ln |x - 1| + C \]

### Final Answer:
\[ \int f(x) \, dx = \frac{2}{3}(x - 1) - \frac{8}{3} \ln |x - 1| + C \]