Improper integral show convergence/divergence $\int_1^{\infty}\frac{3-x-x^2\sin x}{3+x+x^3}\,\mathrm dx$

Question

How do I show the convergence/divergence of this improper integral?

$$\int_1^{\infty}\frac{3-x-x^2\sin x}{3+x+x^3}\,\mathrm dx$$

Answer 1

Using the Intermediate Value Theorem we can show that $3 + x + x^3$ has one root in the interval $(-2,\,0)$. Since it's a monotonically increasing function we conclude that's the only root.

Therefore, in the interval $[1,\,+\infty)$ the function we are integrating is defined and there are no vertical asymptotes. Then observe that $$\frac{3-x-x^2\sin x}{3+x+x^3} \sim_\infty -\frac{\sin x}x$$

Finally, $$-\int_0^\infty\frac{\sin x}x\,\mathrm dx = -\frac\pi2$$ and therefore your integral converges.