<s>[INST] $\int_{0}^{\infty}\frac{x}{x^3+1}dx$ =? 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?
 [/INST] One thing that simplifies the computation is splitting the integral and substituting $y = x^{-1}$ in one part:
$$\begin{align}
\int_0^\infty \frac{x}{x^3+1}\,dx &= \int_0^1\frac{x}{x^3+1}\,dx + \int_1^\infty \frac{x}{x^3+1}\,dx\\
&= \int_0^1 \frac{x}{x^3+1}\,dx + \int_1^0 \frac{y^{-1}}{y^{-3}+1}\,d(y^{-1})\\
&= \int_0^1 \frac{x}{x^3+1}\,dx + \int_0^1 \frac{1}{y^3+1}\,dy\\
&= \int_0^1 \frac{dx}{x^2 -x+1}.
\end{align}$$
This can be seen to be related to an $\arctan$.
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