Monotonicity of Integral Function Defined with Logarithmic and Rational Terms

To determine the monotonicity of the function $g(x)$ on the interval $(1, \infty)$, we differentiate $g(x)$ with respect to $x$ using the Fundamental Theorem of Calculus (Leibniz Rule):

$g'(x) = \left( \frac{2(x - 1)}{x + 1} - \ln x \right) f(x)$

First, let us analyze the sign of $f(x)$ for $x \in (1, \infty)$: $f(x) = (1 - x)^2 \sin^2 x + x^2$ For all $x \in (1, \infty)$, $(1 - x)^2 \sin^2 x \ge 0$ and $x^2 > 1 > 0$. Therefore, $f(x) > 0 \quad \text{for all } x \in (1, \infty)$

Next, define the function $h(x)$ for $x \ge 1$ by: $h(x) = \frac{2(x - 1)}{x + 1} - \ln x$

Evaluating $h(x)$ at $x = 1$: $h(1) = \frac{2(1 - 1)}{1 + 1} - \ln 1 = 0$

Now, we compute the derivative $h'(x)$ for $x > 1$: $h'(x) = \frac{2(x + 1) - 2(x - 1)}{(x + 1)^2} - \frac{1}{x}$ $h'(x) = \frac{4}{(x + 1)^2} - \frac{1}{x} = \frac{4x - (x + 1)^2}{x(x + 1)^2} = \frac{4x - (x^2 + 2x + 1)}{x(x + 1)^2} = -\frac{(x - 1)^2}{x(x + 1)^2}$

For all $x \in (1, \infty)$: $(x - 1)^2 > 0 \quad \text{and} \quad x(x + 1)^2 > 0$ which implies: $h'(x) < 0 \quad \text{for all } x \in (1, \infty)$

Since $h(1) = 0$ and $h'(x) < 0$ for all $x > 1$, the function $h(x)$ is strictly decreasing on $[1, \infty)$. Consequently: $h(x) < h(1) = 0 \quad \text{for all } x \in (1, \infty)$

Now, combining the signs of $h(x)$ and $f(x)$: $g'(x) = h(x) f(x) < 0 \quad \text{for all } x \in (1, \infty)$

Since $g'(x) < 0$ on $(1, \infty)$, the function $g(x)$ is strictly decreasing on $(1, \infty)$.

Therefore, the correct option is B.