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

Question

I want to evaluate the following integral $$ \int_{0}^{\infty} \frac{1}{x^3+x+1}\>dx$$ via Residue theorem. Or, any other methods are welcome!

Actually i just compute this via mathematica, but it seems the command

Integrate[1/(x^3 + x + 1), {x, 0, Infinity}]

gives some terrible output not a compact or simple form.

My first trial is factor $x^3+x+1$ into linear terms but it does not seem easy, I have difficulties for finding poles, $i.e$, finding zeros for $x^3+x+1=0$. And having trouble for finding proper contour.

Answer 1

Note that $x^3+x+1$ has only one real root below per the Cardano’s formula

$$r= \sqrt[3]{\frac{\sqrt{93}-9}{18}}-\sqrt[3]{\frac{\sqrt{93}+9}{18}}= -0.6823 $$

Then, decompose the integrand accordingly to integrate as follows

\begin{align} &\int_0^\infty \frac{1}{x^3+x+1}dx\\ =&\ \int_0^\infty \frac1{(x-r)(x^2+rx-1/r)}dx\\ =&\ \int_0^\infty \frac r{2r +3}\left(-\frac1{x-r}+\frac{x+2r}{x^2+rx-1/r}\right)dx\\ =&\ \frac r{2r +3}\bigg(-\ln\frac{x-r}{\sqrt{ x^2+rx-1/r}} +\frac{ 3r \tan^{-1}\frac{2x+r}{\sqrt{1-3/r} } }{\sqrt{1-3/r}}\bigg)_{0}^{\infty}\\ =&\ \frac {3r}{2r +3}\bigg(\ln(-r)^{\frac12} -\frac{(-r)^{\frac32}}{\sqrt{3-r} } \bigg(\frac\pi2 + \tan^{-1}\frac{(-r)^{\frac32}}{\sqrt{3-r} }\bigg)\bigg) \\ \end{align}