Convergence, absolute convergence of an integral

$\int_{0}^{\infty}\frac{(x^{4}+2017)*(sin x)^{3}}{x^{5}+2018*x^{3}+1}dx$

How do I prove that it converges and that it does not converge absolutely?

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Recall that $\sin^3(x)=\frac34\sin(x)-\frac14\sin(3x)$. Furthermore, $\left|\int_0^L \sin(x)\,dx\right|\le 2$ for all $L$.

Then, note that for sufficiently large $x$, $\frac{x^4+2107}{x^5+2018x^3+1}$ monotonically decreases to $0$.

Abel’s Test (Dirichlet’s Test) for improper integrals guarantees convergence of the integral

\begin{align}\int_0^\infty \frac{(x^4+2107)\sin^3(x)}{x^5+2018x^3+1}\,dx&=\frac34\int_0^\infty \frac{(x^4+2107)\sin(x)}{x^5+2018x^3+1}\,dx-\frac14\int_0^\infty \frac{(x^4+2107)\sin(3x)}{x^5+2018x^3+1}\,dx\end{align}

To show that we have conditional convergence only, we have for $x>16\pi$,

$$\left|\frac{(x^4+2107)\sin^3(x)}{x^5+2018x^3+1}\right|\ge \frac{|\sin^3(x)|}{x}$$

Then, we can write

\begin{align} \int_{16\pi}^{(n+1)\pi}\frac{|\sin^3(x)|}{x}\,dx&=\sum_{k=16}^n\int_{k\pi}^{(k+1)\pi}\frac{|\sin^3(x)|}{x}\,dx\\\\ &\ge \sum_{k=16}^n \frac{1}{(k+1)\pi}\int_0^\pi \sin^3(x)\,dx\\\\ &=\sum_{k=16}^n \frac{4}{3(k+1)\pi}\tag 1 \end{align}

Since, the series in $(1)$ diverges by comparison with the harmonic series, then the integral of interest does not absolutely converge.