Whereabouts of $g$ .

Let us examine the derivative of $g(x)$ :

$g'(x) = [2(\frac{x-1}{x+1}) - \ln(x)] \cdot [(1-x)^{2} \sin^{2}(x) + x^2] = p(x)f(x)$ .

Upon observation, $f(x) \ge 0$ for all $x \in \mathbb{R}$ with one root at $x=0$ , and it's strictly positive over $(1, \infty).$ Taking the derivatives of $p(x)$ yield:

$p'(x) = \frac{4}{(x+1)^2} - \frac{1}{x}$ and $p''(x) = -\frac{8}{(x+1)^3} + \frac{1}{x^2}$

with $p'(1) = p''(1) = 0, p'(x) < 0$ (for $x>1$ ), which makes $p(x)$ strictly negative over $(1,\infty)$ . Altogether, $g'(x) = Negative \times Positive = Negative \Rightarrow g(x)$ is a decreasing function over $(1,\infty)$ .