Answer
Convergent.
Work Step by Step
$ I_1 = \int\limits_{0}^{\infty}\frac{x}{x^3+1}\, dx$
Let $I_2 = \int\limits_{0}^{\infty}\frac{x+1}{x^3+1}\, dx$,
We know that $I_2\geq I_1$, Because the denominator is bigger and it's always positive.
Let's solve for $I_2$
$I_2 = \int\limits_{0}^{\infty}\frac{x+1}{(x+1)(x^2-x+1)}\, dx$
$I_2 = \int\limits_{0}^{\infty}\frac{1}{x^2-x+1}\, dx$
By completing the square we get:
$I_2= \int\limits_{0}^{\infty}\frac{1}{x^2-x+\frac{1}{4}+\frac{3}{4}}\, dx=\int\limits_{0}^{\infty}\frac{1}{(x-\frac{1}{2})^2+\frac{3}{4}}\, dx$, Now multiplying top and bottom by $\frac{4}{3}$
$I_2 = \frac{4}{3}\int\limits_{0}^{\infty}\frac{1}{(\frac{2x-1}{\sqrt{3}})^2+1}\, dx$
Using the fact: $\int\frac{1}{(Au+B)^2+1}\, du= \frac{tan^{-1}(Au+B)}{A}+C$
$I_2 = \frac{4}{3}\int\limits_{0}^{\infty}\frac{1}{(\frac{2x-1}{\sqrt{3}})^2+1}\, dx= \frac{4}{3}\frac{tan^{-1}(\frac{2x-1}{\sqrt{3}})}{\frac{2}{\sqrt{3}}}\Biggr|_0^{\infty}= \frac{4\times\sqrt3}{2\times3}tan^{-1}(\frac{2x-1}{\sqrt{3}})\Biggr|_0^{\infty}$
$I_2=\frac{2}{\sqrt3}tan^{-1}(\frac{2x-1}{\sqrt{3}})\Biggr|_0^{\infty}= \frac{2}{\sqrt3}(\frac{\pi}{2}-\frac{-\pi}{6})=\frac{2}{\sqrt3}(\frac{2\pi}{3})=\frac{4\pi}{3\sqrt3}.$
$I_2$ $And$ $I_1$ $Converges$.
