If `f(x) = log ((1+x)/(1-x))`, where `-1 lt x lt 1` then `f((3x+x^(3))/(1+3x^(2))) - f((2x)/(1+x^(2)))` is equal to

Options

A. `[f(x)]^(3)`
B. `[f(x)]^(2)`
C. `-f(x)`
D. f(x)

Answer

To solve the problem, we need to evaluate the expression \( f\left(\frac{3x + x^3}{1 + 3x^2}\right) - f\left(\frac{2x}{1 + x^2}\right) \) where \( f(x) = \log\left(\frac{1+x}{1-x}\right) \).

### Step-by-Step Solution:

1. **Identify the Function**:
We have \( f(x) = \log\left(\frac{1+x}{1-x}\right) \).

2. **Substitute the First Expression**:
Let \( a = \frac{3x + x^3}{1 + 3x^2} \).
Then, we need to find \( f(a) = \log\left(\frac{1 + a}{1 - a}\right) \).

3. **Calculate \( 1 + a \) and \( 1 - a \)**:
- \( 1 + a = 1 + \frac{3x + x^3}{1 + 3x^2} = \frac{(1 + 3x^2) + (3x + x^3)}{1 + 3x^2} = \frac{1 + 3x + 3x^2 + x^3}{1 + 3x^2} \)
- \( 1 - a = 1 - \frac{3x + x^3}{1 + 3x^2} = \frac{(1 + 3x^2) - (3x + x^3)}{1 + 3x^2} = \frac{1 - 3x + 3x^2 - x^3}{1 + 3x^2} \)

4. **Combine the Expressions**:
Now, we can express \( f(a) \):
\[ f(a) = \log\left(\frac{1 + a}{1 - a}\right) = \log\left(\frac{1 + 3x + 3x^2 + x^3}{1 - 3x + 3x^2 - x^3}\right) \]

5. **Substitute the Second Expression**:
Let \( b = \frac{2x}{1 + x^2} \).
Then, we need to find \( f(b) = \log\left(\frac{1 + b}{1 - b}\right) \).

6. **Calculate \( 1 + b \) and \( 1 - b \)**:
- \( 1 + b = 1 + \frac{2x}{1 + x^2} = \frac{(1 + x^2) + 2x}{1 + x^2} = \frac{1 + 2x + x^2}{1 + x^2} \)
- \( 1 - b = 1 - \frac{2x}{1 + x^2} = \frac{(1 + x^2) - 2x}{1 + x^2} = \frac{1 - 2x + x^2}{1 + x^2} \)

7. **Combine the Expressions**:
Now, we can express \( f(b) \):
\[ f(b) = \log\left(\frac{1 + b}{1 - b}\right) = \log\left(\frac{1 + 2x + x^2}{1 - 2x + x^2}\right) \]

8. **Subtract the Two Functions**:
Now we need to find:
\[ f(a) - f(b) = \log\left(\frac{1 + 3x + 3x^2 + x^3}{1 - 3x + 3x^2 - x^3}\right) - \log\left(\frac{1 + 2x + x^2}{1 - 2x + x^2}\right) \]
This simplifies to:
\[ = \log\left(\frac{(1 + 3x + 3x^2 + x^3)(1 - 2x + x^2)}{(1 - 3x + 3x^2 - x^3)(1 + 2x + x^2)}\right) \]

9. **Final Result**:
After simplification, we find that:
\[ f\left(\frac{3x + x^3}{1 + 3x^2}\right) - f\left(\frac{2x}{1 + x^2}\right) = 0 \]

### Conclusion:
Thus, the answer is \( 0 \).