To determine the nature of the function $g(x)$, we need to analyze the sign of its derivative, $g'(x)$.

Step 1: Find the derivative of $g(x)$

The function $g(x)$ is defined as: $g(x) = \int\limits_1^x {\left( {{{2(t - 1)} \over {t + 1}} - \ln t} \right)f(t)dt}$ Using the Leibniz rule (Fundamental Theorem of Calculus, Part 1), we can find the derivative of $g(x)$ with respect to $x$: $g'(x) = \frac{d}{dx} \int\limits_1^x {\left( {{{2(t - 1)} \over {t + 1}} - \ln t} \right)f(t)dt}$ $g'(x) = \left( {{{2(x - 1)} \over {x + 1}} - \ln x} \right)f(x)$

Step 2: Analyze the sign of $f(x)$

The function $f(x)$ is given by: $f(x) = (1 - x)^2 \sin^2 x + x^2$ We need to determine the sign of $f(x)$ for $x \in (1, \infty)$.

• The term $(1 - x)^2$ is always non-negative. For $x > 1$, $(1-x)^2 > 0$.
• The term $\sin^2 x$ is always non-negative, i.e., $0 \le \sin^2 x \le 1$.
• The term $x^2$ is always non-negative. For $x \in (1, \infty)$, we have $x^2 > 1$.

The first term, $(1 - x)^2 \sin^2 x$, is greater than or equal to zero. The second term, $x^2$, is strictly positive for $x \in (1, \infty)$. Therefore, the sum $f(x) = (1 - x)^2 \sin^2 x + x^2$ is strictly positive for all $x \in (1, \infty)$. $f(x) > 0 \quad \text{for all } x \in (1, \infty)$

Step 3: Analyze the sign of the other factor

Since $f(x) > 0$, the sign of $g'(x)$ is determined by the sign of the other factor. Let's define a new function $h(x)$: $h(x) = \frac{2(x - 1)}{x + 1} - \ln x \quad \text{for } x \in (1, \infty)$ To find the sign of $h(x)$, we analyze its derivative, $h'(x)$.

First, find the derivative of the fractional term using the quotient rule: $\frac{d}{dx} \left( \frac{2(x - 1)}{x + 1} \right) = 2 \frac{(1)(x+1) - (x-1)(1)}{(x+1)^2} = 2 \frac{x+1-x+1}{(x+1)^2} = \frac{4}{(x+1)^2}$ Now, find the derivative of $h(x)$: $h'(x) = \frac{4}{(x + 1)^2} - \frac{1}{x}$ To determine the sign of $h'(x)$, we combine the terms: $h'(x) = \frac{4x - (x+1)^2}{x(x+1)^2} = \frac{4x - (x^2 + 2x + 1)}{x(x+1)^2}$ $h'(x) = \frac{-x^2 + 2x - 1}{x(x+1)^2} = \frac{-(x^2 - 2x + 1)}{x(x+1)^2}$ $h'(x) = \frac{-(x - 1)^2}{x(x + 1)^2}$ For $x \in (1, \infty)$:

• The numerator $-(x - 1)^2$ is strictly negative (since $x \ne 1$).
• The denominator $x(x + 1)^2$ is strictly positive. Therefore, $h'(x) < 0$ for all $x \in (1, \infty)$.

This means that the function $h(x)$ is strictly decreasing on the interval $(1, \infty)$.

Step 4: Determine the sign of $h(x)$

Since $h(x)$ is strictly decreasing on $(1, \infty)$, its value for any $x > 1$ will be less than its value at $x=1$. Let's evaluate $h(x)$ at $x=1$: $h(1) = \frac{2(1 - 1)}{1 + 1} - \ln(1) = \frac{0}{2} - 0 = 0$ Since $h(x)$ is strictly decreasing for $x > 1$ and $h(1)=0$, it follows that: $h(x) < h(1) = 0 \quad \text{for all } x \in (1, \infty)$ So, $h(x)$ is negative on the interval $(1, \infty)$.

Step 5: Conclude the sign of $g'(x)$ and the monotonicity of $g(x)$

We have $g'(x) = h(x) \cdot f(x)$.

• For $x \in (1, \infty)$, we found $f(x) > 0$.
• For $x \in (1, \infty)$, we found $h(x) < 0$. Therefore, the product $g'(x) = (\text{negative}) \times (\text{positive})$ is negative. $g'(x) < 0 \quad \text{for all } x \in (1, \infty)$ Since the derivative of $g(x)$ is negative throughout the interval $(1, \infty)$, the function $g(x)$ is decreasing on $(1, \infty)$.

This corresponds to option B.