$\int_{0}^{\infty}\frac{x}{x^3+1}dx$ =? 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?

If $I=\int_{0}^{\infty}\frac{x}{x^3+1}dx$ then $2I=\int_{0}^{\infty}\frac{2x-x^2+x^2}{x^3+1}dx$=$\int_{0}^{\infty}\frac{2x-x^2}{x^3+1}dx$+$\int_{0}^{\infty}\frac{x^2}{x^3+1}dx.$
 The secod itegral is easy and can be solved by substituting $u=x^3+1$. For the first write  $$\int_{0}^{\infty}\frac{2x-x^2}{x^3+1}dx=\int_{0}^{\infty}\frac{(x+1)-(x^2-x+1)}{(x+1)(x^2-x+1)}dx=\int_{0}^{\infty}\frac{1}{x^2-x+1}dx-\int_{0}^{\infty}\frac{1}{x+1}dx$$
but for the $\int_{0}^{\infty}\frac{1}{x^2-x+1}dx$=$\int_{0}^{\infty}\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}dx$ which is of the form $\int \frac{1}{u^2+a^2}du=\frac{1}{a}\arctan(\frac{u}{a})+c.$
