# Does this integral converge? $\int_2^\infty \frac{x \sin^2 x}{x^3-1}\ \mathsf dx$

Does $$\int_2^\infty \frac{x \sin^2 x}{x^3-1} dx$$ converge or diverge? I think this is a comparison test but I'm not sure what integral I should compare it with. Thanks

Edit: The integral used to be from $1$ to $\infty$, now it is from $2$ to $\infty$. So the badness of our function has become irrelevant.
If $x\ge 2$, then the top is positive and $\le x$, and the bottom is greater than $x^3/2$. So our function is less than $\frac{2}{x^2}$. Since $\int_2^\infty \frac{2}{x^2}\,dx$ converges, so does our integral.
Old answer: Note that $x^3-1=(x-1)(x^2+x+1)$. If $x$ is reasonably near to $1$ but greater than $1$, our function is greater than $$\frac{\sin^2(1)}{4(x-1)}.$$ But $\int_1^{1+\delta}\frac{dx}{x-1}$ diverges for any positive $\delta$, so our integral diverges.