About the convergence of the particular type of improper integral

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I want to know about the convergence of the following integral :

$$ \int_1^{+\infty}\sin(x^3+x)dx $$

I have tried a few things here and there but I am nowhere close to prove the convergence or divergence.

I was also thinking about the following way :

If $I\subset[1,\infty)$ is any compact interval, then, if possible, we can show that $\int_{I}\sin(x^3+x)dx$ is uniformly bounded.

But I am still unable to prove this.

I would be grateful for any hint or help.

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HINT

$$\int_1^{+\infty}\sin(x^3+x)dx=\int_1^{+\infty}\frac{1}{3x^2+1}\cdot(3x^2+1)\sin(x^3+x)dx$$

Now, integrate by parts with $dv=(3x^2+1)\sin(x^3+x)$

Can you finish?