$\int_{0}^{\infty}\frac{x}{x^3+1}dx$ =? So guys I have this improper integral $\int_{0}^{\infty}\frac{x}{x^3+1}dx$. I checked that it converges by $ \int_{0}^{1}\frac{x}{x^3+1}dx + \int_{1}^{\infty}\frac{x}{x^3+1}dx $ and using the $\frac{c}{(x-a)^\lambda}  $ and $\frac{1}{x^\lambda} $ criteria. But for finding the value I have trouble.I tried $ \int_{0}^{\infty}\frac{x}{(x+1)(x^2-x+1)}dx$ and $\int_{0}^{\frac{\pi}{2}}\frac{tg\alpha}{tg\alpha^3+1}dx$ but it gets me nowhere. Any ideas on solving it?

Continuing Daniel Fischer's answer, we have:
$$\int_{0}^{1}\frac{dx}{x^2-x+1}=\int_{-1/2}^{1/2}\frac{dx}{x^2+3/4}=\frac{\sqrt{3}}{2}\cdot\frac{4}{3}\int_{-1/\sqrt{3}}^{1/\sqrt{3}}\frac{dx}{x^2+1}=\frac{4\sqrt{3}}{3}\arctan\frac{1}{\sqrt{3}}=\frac{2\pi}{3\sqrt{3}}.$$
