Evaluation of definite integral using residue theorem $$ \int^{+\infty}_{-\infty} \frac{x-1}{x^3-1} dx$$
I need to evaluate the above integral . 
My idea is to consider the same integral but with the $x$'s as $z$'s, over the complex plane, have a closed contour integral over $\gamma$, and then use the residue theorem. i.e. consider: 
$$ \int^{+\infty}_{-\infty} \frac{z-1}{z^3-1} dz$$
I'm stuck on how to formulate $\gamma$ though.
I know this has 3 poles: at 
$z=1$, $z= \frac{-1}{2} + i\frac{\sqrt3}{2}$ and $z= \frac{-1}{2} - i\frac{\sqrt3}{2}$
How do I use this to divide up gamma over contours to which I can then use the residue theorem? And then do I have to either evaluate directly or apply the ML inequality to each individual contour?

Note that $\frac{x - 1}{x^3 - 1} = \frac{1}{x^2 + x + 1}$. So the integral is
$$
\int_{-\infty}^{+\infty} \frac{1}{x^2 + x + 1} \, dx.
$$
First take the integral from $-a$ to $a$ and then take the limit after. Just complete the square in the denominator and use substitution to get an integral in terms of $\arctan$.
$$
\int_{-a}^{a} \frac{1}{(x + \frac{1}{2})^2 + \frac{3}{4}} \, dx \\
=\frac{4}{3}\int_{-a}^{a} \frac{1}{\left(\frac{2}{\sqrt{3}}(x + \frac{1}{2})\right)^2 + 1} \, dx.
$$
Using the substitution $u = \frac{2}{\sqrt{3}}(x + \frac{1}{2})$, we have $dx = \frac{\sqrt{3}}{2}du$ and so the indefinite integral is
$$
\frac{4}{3}\frac{\sqrt{3}}{2}\int \frac{1}{u^2 + 1} \, dx \\
=\frac{2}{\sqrt{3}}\arctan(u) = \frac{2}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}(2x + 1)\right).
$$
Thus, the definite integral is
$$
\left.\frac{2}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}(2x + 1)\right)\right|_{-a}^a = \frac{2}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}(2a + 1)\right) - \frac{2}{\sqrt{3}}\arctan\left(\frac{1}{\sqrt{3}}(-2a + 1)\right).
$$
As $a \to \infty$, we have $\arctan\left(\frac{1}{\sqrt{3}}(2a + 1)\right) \to \frac{\pi}{2}$ and $\arctan\left(\frac{1}{\sqrt{3}}(-2a + 1)\right) \to -\frac{\pi}{2}$. Hence the integral is
$$
\int_{-\infty}^{+\infty} \frac{x - 1}{x^3 - 1} \, dx= \frac{2}{\sqrt{3}} \cdot \frac{\pi}{2} + \frac{2}{\sqrt{3}} \cdot \frac{\pi}{2} = \frac{2\pi}{\sqrt{3}}.
$$
