How can I establish the convergence (or divergence) of $ \int_{0}^{\infty} \frac{x^2 \ln(\sin^2(x))}{(1+x^2)^2} \mathrm{d}x $ I want to find out wether the following  integral converges
$ \displaystyle \int_{0}^{\infty} \frac{x^2 \ln(\sin^2(x))}{(1+x^2)^2} \mathrm{d}x \tag{1} $
The integral above converges iff the following integral converges.
$ \displaystyle \int_{0}^{\infty} \frac{x^2 \ln \left( \frac{\sin^2(x)}{x^2} \right) }{(1+x^2)^2} \mathrm{d}x \tag{2} $
Since we have $ \frac{\sin^2(x)}{x^2} <1 $ for all $ x \in \mathbb{R} $, we can say that
$ \displaystyle \ln \left( \frac{\sin^2(x)}{x^2} \right) =- \sum_{k=1}^{\infty} \frac{1}{k} \left(1- \frac{\sin^2(x)}{x^2} \right)^k \tag*{} $
At this point, if we can somehow prove that
$\displaystyle \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2}  \left(1- \frac{\sin^2(x)}{x^2} \right)^k < \frac{1}{k^{\epsilon}} \tag*{} $
For some $ \epsilon >0 $, Then the claim of convergence of the original integral will follow, but I can't seem to find a way to do so.
P.S. Wolframalpha has given up on providing a numerical estimate for the integral in (1), but says that the integral in (2) diverges, which is kind of puzzling.

Note that $x\to \ln(\sin^2(x))\leq 0$ and it is periodic of period $\pi$. Notice that
$$\int_0^{\pi}\ln(\sin^2(x))\,dx=-2\pi\ln(2).$$
(see for example Complex integration with trigonometric and logarithm? ).
Moreover $x\to\frac{x^2 }{(1+x^2)^2}\geq 0$, it is increasing in $[0,1]$ and decreasing in $[1,+\infty)$. Hence
$$\begin{align}\int_{0}^{\infty} &\frac{x^2 |\ln(\sin^2(x))|}{(1+x^2)^2}\, dx=
\sum_{k=0}^{\infty}\int_{k\pi}^{(k+1)\pi}\frac{x^2 |\ln(\sin^2(x))|}{(1+x^2)^2}\, dx\\
&\leq \frac{1}{4}\int_{0}^{\pi}|\ln(\sin^2(x))| dx+\sum_{k=1}^{\infty}\frac{(k\pi)^2 }{(1+(k\pi)^2)^2}\int_{k\pi}^{(k+1)\pi}|\ln(\sin^2(x))|\, dx\\
&\leq \frac{\pi\ln(2)}{2}+2\pi\ln(2)\sum_{k=1}^{\infty}\frac{(k\pi)^2 }{(k\pi)^4}\\
&= \frac{\pi\ln(2)}{2}+\frac{2\ln(2)}{\pi}\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi\ln(2)}{2}+\frac{\pi\ln(2)}{3}=\frac{5\pi\ln(2)}{6}
<+\infty\end{align}$$
Hence the integral is convergent.
