Judge the convergence of the $\int_{0}^{\infty}\frac{x\ dx}{1+x^4\sin^2x}$ 
Judege if the improper integral is convergence $\int_{0}^{\infty}\frac{x\ dx}{1+x^4\sin^2x}$.

My work:
I want to change it into a series: the series=$\sum_{0}^{\infty}\int_{k\pi}^{(k+1)\pi}\frac{x\ dx}{1+x^4\sin^2x}$, and then use the substitution, $\int_{k\pi}^{(k+1)\pi}\frac{x\ dx}{1+x^4\sin^2x}=\int_{0}^{\pi}\frac{x+k\pi}{1+(x+k\pi)^4\sin^2(x)}dx$. And change the series into two series $$\sum_{k=0}^{\infty}\int_{0}^{\pi}\frac{x}{1+(x+k\pi)^4\sin^2(x)}dx+\sum_{k=0}^{\infty}\int_{0}^{\pi}\frac{k\pi}{1+(x+k\pi)^4\sin^2(x)}dx$$ but I do not how to do next.
Thank you for any help.
 A: Informal outline: On each interval $(n\pi - n^{-2}, n\pi + n^{-2})$, we have that $\sin^2 x < n^{-4}$, so $1 + x^4 \sin^2 x$ is bounded above on the union of these intervals (with the $\limsup$ being $1 + \pi^4$).  Therefore, $f(x) = \Omega(x)$ when restricted to the union of these intervals, and the integral of $f(x)$ over the $n$th interval is $\Omega(\frac{1}{n})$.  The sum of these lower bounds of contributions is divergent, and $f(x) > 0$ everywhere on $(0, \infty)$; so we can conclude that $\int_0^\infty f(x)\,dx$ must also be divergent.
A: Use the fact:
$$0\leq \sin x\leq x,\ 0\leq x\leq \frac\pi2.$$
$$\int_{0}^{\pi}\frac{k\pi}{1+(x+k\pi)^4\sin^2(x)}dx>\int_{0}^{\pi}\frac{k\pi}{1+(\pi+k\pi)^4\sin^2(x)}dx\\>
\int_{0}^{\frac\pi2}\frac{k\pi}{1+(\pi+k\pi)^4\sin^2(x)}dx
>\int_{0}^{\frac\pi2}\frac{k\pi}{1+(\pi+k\pi)^4x^2}dx\\
=\frac{k\pi}{(k+1)^2\pi^2}\int_{0}^{\frac\pi2}\frac{d((k+1)^2\pi^2x)}{1+(\pi+k\pi)^4x^2}
=\frac{k\pi}{(k+1)^2\pi^2}\arctan((k+1)^2\pi^2x)\Bigg|_{0}^{\frac\pi2}\\
>\frac{k}{(k+1)^2\pi}\sim\frac{1}{k\pi}.$$
