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?
 A: 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.
