Evaluating the improper integral $\int_0^{\infty}\frac{dx}{1+x^3}$ 
Evaluate $$\int_{0}^{\infty}\frac{dx}{1+x^3}.$$

I tried integration by partial fraction. My work is below:
$$\int_{0}^{\infty}\frac{dx}{1+x^3}=\frac{1}{3} \int_{0}^{\infty}\frac{1}{x+1}+\frac{1}{3} \int_{0}^{\infty}\frac{2-x}{x^2-x+1}.$$
It seems that the result would go to infinity. But the answer is $\dfrac{2\pi}{3\sqrt3}$.
Did I do something wrong? 
 A: Hint. You may first integrate over the finite set $[0,M]$ then let $M \to \infty$:
$$\int_{0}^M\frac{dx}{1+x^3}=\frac{1}{3} \int_{0}^M\frac{1}{x+1}\:dx+\frac{1}{3} \int_{0}^M\frac{2-x}{x^2-x+1}\:dx,$$ you will see that the $\log$ terms cancel.

Edit. Let $M>0$. We have $$\frac{1}{3} \int_{0}^M\frac{1}{x+1}\:dx=\frac{1}{3}\log (M+1)$$
$$\frac{1}{3} \int_{0}^M\frac{2-x}{x^2-x+1}\:dx=-\frac{1}{6}\log (M^2-M+1)-\frac1{\sqrt3}\arctan\left(\frac{2 M-1}{\sqrt{3}}\right)+\frac{\pi}{6\sqrt3}.$$
You may observe that, as $M \to \infty$,
$$
\frac{1}{3}\log (M+1)-\frac{1}{6}\log (M^2-M+1)=\frac{1}{6}\log \left(\frac{1+2/M+1/M^2}{1-1/M+1/M^2} \right) \to 0.
$$
A: \begin{align}
\int_{0}^{\infty}\frac{1}{1+x^3}dx&\overset{x\to\frac1x}=\int_{0}^{\infty}\frac{x}{1+x^3}dx
= \frac12\int_{0}^{\infty}\frac{1+x}{1+x^3}dx\\
& = \frac12\int_{0}^{\infty}\frac{1}{x^2-x+1}dx
= \frac12\int_{0}^{\infty}\frac{d(x-\frac12)}{(x-\frac12)^2+\frac34}\\
&=\frac1{\sqrt3}\tan^{-1}\frac{2x-1}{\sqrt3}\bigg|_0^\infty= \frac1{\sqrt3}\left(\frac\pi2- (-\frac\pi6)\right)=\frac{2\pi}{3\sqrt3}
\end{align}
