Residue Theorem Integral

Studying for qualifying exams, I came across the following problem: using complex analysis, compute
$$int_{-infty}^{infty}frac{x^2sin(pi x)}{x^3-1}dx
$$
I decided to use the integrand $f(z)=frac{z^2e^{ipi z}}{z^3-1}$ (the goal is to take the imaginary part in the end), and I found that $f$ has a removable singularity at $1$. Now, it seems to me that $|f(Re^{itheta})|=O(e^R/R)$, so a semicircle countour won’t work. I also tried to use a rectangular contour (with height $2pi$) and the sides do vanish as $Rtoinfty$, but since there are quadratic terms in the integrand, I am not able to get a simple result and conclude via the residue theorem. Any ideas?

Edit: With help from the comment, I think I have a solution. Taking $f(z)$ as before, we see that $f(z)=e^{ipi z}cdot g(z)$, where $g(z)=frac{z^2}{z^3-1}$, and since $|g(Re^{itheta})|leq frac{C}{R}$, we invoke Jordan’s lemma to say that $lim_{Rtoinfty}int_{Gamma_R}f(z)=0$, where $Gamma_R$ is the upper semi-circle of radius $R$ centered at $0$. Therefore, we get:
$$int_{-infty}^{infty}frac{xe^{ipi x}}{x^3-1}=2pi i Res_{e^{2pi i/3}}f
$$
Now,
$$Res_{e^{2pi i/3}}f=lim_{zto e^{2pi i/3}}frac{ze^{ipi z}}{(z-1)(z-e^{4ipi /3})}=frac{e^{2pi i/3}exp{ipi e^{2pi i/3}}}{(e^{2pi i/3}-1)(e^{2pi i/3}-e^{4pi i/3})}
$$
This will simplify (using the fact that the roots of unity sum to $0$) to $exp{ipi( e^{2pi i/3}-2/3)}$
Now, $Im{2pi i exp{ipi( e^{2pi i/3}-2/3)}}=2pi e^{-pi sin(2pi/3)}cos[cos(2pi/3)-2/3]$. I don’t think this should be the answer, but I am unable to figure out where I went wrong.