Let $$f$$ be a real valued function, defined on $$R - \{-1, 1\}$$ and given by $$f(x) = 3\log_e\left|\frac{x-1}{x+1}\right| - \frac{2}{x-1}$$. Then in which of the following intervals, function $$f(x)$$ is increasing?

We have $$f(x) = 3\ln\left|\frac{x-1}{x+1}\right| - \frac{2}{x-1}$$, defined on $$\mathbb{R} \setminus \{-1, 1\}$$.

Differentiating: $$f'(x) = 3\left(\frac{1}{x-1} - \frac{1}{x+1}\right) + \frac{2}{(x-1)^2} = \frac{3}{x-1} - \frac{3}{x+1} + \frac{2}{(x-1)^2}$$.

Taking the common denominator $$(x-1)^2(x+1)$$:

$$f'(x) = \frac{3(x-1)(x+1) - 3(x-1)^2 + 2(x+1)}{(x-1)^2(x+1)}$$

Expanding the numerator: $$3(x^2 - 1) - 3(x^2 - 2x + 1) + 2(x + 1) = 3x^2 - 3 - 3x^2 + 6x - 3 + 2x + 2 = 8x - 4 = 4(2x - 1)$$.

So $$f'(x) = \frac{4(2x-1)}{(x-1)^2(x+1)}$$. Since $$(x-1)^2 > 0$$ for $$x \neq 1$$, the sign of $$f'(x)$$ depends on $$\frac{2x-1}{x+1}$$.

$$f'(x) \geq 0$$ when $$\frac{2x-1}{x+1} \geq 0$$, which happens when both factors are non-negative or both non-positive:

Case 1: $$2x - 1 \geq 0$$ and $$x + 1 > 0$$, i.e., $$x \geq \frac{1}{2}$$ (with $$x \neq 1$$).

Case 2: $$2x - 1 \leq 0$$ and $$x + 1 < 0$$, i.e., $$x < -1$$.

Therefore, $$f(x)$$ is increasing on $$(-\infty, -1) \cup \left[\frac{1}{2}, \infty\right) \setminus \{1\}$$.