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 \( f(x) \) defined by the equation \( f\left(\frac{3x-4}{3x+4}\right) = x + 2 \).

### Step-by-Step Solution:

1. **Substitution**:
Let \( a = \frac{3x-4}{3x+4} \). We will express \( x \) in terms of \( a \).

Rearranging gives:
\[ a(3x + 4) = 3x - 4 \]
\[ 3ax + 4a = 3x - 4 \]
\[ 3ax - 3x = -4 - 4a \]
\[ x(3a - 3) = -4(1 + a) \]
\[ x = \frac{-4(1 + a)}{3(a - 1)} \]

**Hint**: Use algebraic manipulation to isolate \( x \).

2. **Finding \( x + 2 \)**:
Now, we need to find \( x + 2 \):
\[ x + 2 = \frac{-4(1 + a)}{3(a - 1)} + 2 \]
Converting 2 to a fraction with the same denominator:
\[ x + 2 = \frac{-4(1 + a) + 6(a - 1)}{3(a - 1)} \]
Simplifying the numerator:
\[ = \frac{-4 - 4a + 6a - 6}{3(a - 1)} = \frac{2a - 10}{3(a - 1)} \]

**Hint**: Remember to combine fractions by finding a common denominator.

3. **Expressing \( f(a) \)**:
From the original function, we have:
\[ f(a) = x + 2 = \frac{2a - 10}{3(a - 1)} \]

**Hint**: Substitute back the expression for \( x + 2 \) into \( f(a) \).

4. **Finding \( f(x) \)**:
Now, we can express \( f(x) \) as:
\[ f(x) = \frac{2x - 10}{3(x - 1)} \]

**Hint**: Ensure the variable in \( f \) matches the variable in the expression.

5. **Integrating \( f(x) \)**:
We need to compute the integral:
\[ \int f(x) \, dx = \int \frac{2x - 10}{3(x - 1)} \, dx \]
This can be split into two parts:
\[ = \frac{1}{3} \int (2 - \frac{8}{x - 1}) \, dx \]
Now, integrating term by term:
\[ = \frac{1}{3} \left( 2x - 8 \ln |x - 1| \right) + C \]
Simplifying gives:
\[ = \frac{2x}{3} - \frac{8}{3} \ln |x - 1| + C \]

**Hint**: Use the linearity of integration to separate the terms.

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