How should I solve this improper integral ? $$\int_{2}^{\infty}\frac{x}{x^3-2\sin x}dx$$ I tried symbolab integral calculator but it was unable to solve. Any suggestions would be appreciated! Thank you
2026-04-01 07:19:07.1775027947
How to solve this improper integral $\int_{2}^{\infty}\frac{x}{x^3-2\sin x}dx$
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We know that $-1\leq \sin(x)\leq 1$ thus $\int_2^{\infty} \frac{xdx}{x^3-2\sin(x)} < \int_2^{\infty} \frac{xdx}{x^3-2}$ the latter integral is comparable $\frac{1}{x^2}$ when $x$ becomes very large. You can find the integrand of $\int \frac{xdx}{x^3-2}$ manually thus the original integral converges.