Compute $\int_0^\infty \frac{dx}{1+x^3}$ Problem
Compute $$\displaystyle \int_0^\infty \frac{dx}{1+x^3}.$$
Solution
I do partial fractions
$$\frac{1}{x^3+1}= \frac{2-x}{3 \left( x^{2}-x+1 \right)}+\frac{1}{3 \left( x+1 \right)}.$$
But we could simplify the left one $$\frac{2-x}{3\left( x^{2}-x+1 \right)} = \frac{2}{3}\cdot \frac{1}{x^{2}-x+1}-\frac{x}{x^{2}-x+1}$$          
From here, I do this see images.
But I find the wrong primitive functions. Why am I wrong, and how do I find the correct one?
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$\ds{\int_{0}^{\infty}{\dd x \over 1 + x^{3}}:\ {\large ?}}$

\begin{align}
&\color{#c00000}{\int_{0}^{\infty}{\dd x \over 1 + x^{3}}}
=\int_{0}^{\infty}\int_{0}^{\infty}\expo{-\pars{1 + x^{3}}t}\,\dd t\,\dd x
=\int_{0}^{\infty}\expo{-t}\ \overbrace{\int_{0}^{\infty}\expo{-t\,x^{3}}\,\dd x}
^{\ds{tx^{3}\equiv\xi\imp x=t^{-1/3}\xi^{1/3}}}\ \,\dd t
\\[3mm]&=\int_{0}^{\infty}\expo{-t}
\int_{0}^{\infty}\expo{-\xi}\,t^{-1/3}\,{1 \over 3}\,\xi^{-2/3}\,\dd\xi\,\dd t
\\[3mm]&={1 \over 3}\pars{\int_{0}^{\infty}t^{-1/3}\expo{-t}\,\dd t}
\pars{\int_{0}^{\infty}\xi^{-2/3}\expo{-\xi}\,\dd\xi}
={1 \over 3}\,\Gamma\pars{2 \over 3}\Gamma\pars{1 \over 3}
\end{align}
  where $\ds{\Gamma\pars{z}}$ is the Gamma Function ${\bf\mbox{6.1.1}}$.

With Euler Reflection Formula
${\bf\mbox{6.1.17}}$:
\begin{align}
&\color{#c00000}{\int_{0}^{\infty}{\dd x \over 1 + x^{3}}}
={1 \over 3}\,{\pi \over \sin\pars{\pi/3}}
={1 \over 3}\,{\pi \over \root{3}/2}
\end{align}

$$\color{#00f}{\large%
\int_{0}^{\infty}{\dd x \over 1 + x^{3}} = {2\root{3} \over 9}\,\pi}
\approx 1.2092
$$

A: Since
$$
\frac{1}{x^3+1}=\frac{1}{(x+1)(x^2-x+1)},
$$
we can find some $a,b,c \in \mathbb{R}$ such that
$$
\frac{1}{x^3+1}=\frac{a}{x+1}+\frac{bx+c}{x^2-x+1}.
$$
A simple computation shows that
$$
a=-b=\frac{c}{2}=\frac13,
$$
i.e.
\begin{eqnarray}
\frac{1}{x^3+1}&=&\frac13\cdot\frac{1}{x+1}-\frac13\cdot\frac{x-2}{x^2-x+1}\\
&=&\frac13\cdot\frac{1}{x+1}-\frac16\cdot\frac{2x-1}{x^2-x+1}+\frac12\cdot\frac{1}{x^2-x+1}\\
&=&\frac13\cdot\frac{1}{x+1}-\frac16\cdot\frac{2x-1}{x^2-x+1}+\frac12\cdot\frac{1}{\left(x-\frac12\right)^2+\frac34}.
\end{eqnarray}
It follows that
\begin{eqnarray}
F(r)&=&\int_0^r\frac{1}{x^3+1}\,dx=\frac13\ln(1+r)-\frac16\ln(r^2-r+1)+\frac{1}{\sqrt{3}}\arctan\frac{2r-1}{\sqrt{3}}+\frac{1}{\sqrt{3}}\arctan\frac{1}{\sqrt{3}}\\
&=&\frac16\ln\frac{(r+1)^2}{r^2-r+1}+\frac{1}{\sqrt{3}}\arctan\frac{2r-1}{\sqrt{3}}+\frac{\pi}{6\sqrt{3}}\\
&=&\frac16\ln\frac{r^2+2r+1}{r^2-r+1}+\frac{1}{\sqrt{3}}\arctan\frac{2r-1}{\sqrt{3}}+\frac{\pi}{6\sqrt{3}}.
\end{eqnarray}
Thus
$$
\int_0^\infty\frac{1}{x^3+1}\,dx=\lim_{r\to\infty}F(r)=\frac{\pi}{2\sqrt{3}}+\frac{\pi}{6\sqrt{3}}=\frac{2\pi}{3\sqrt{3}}.
$$
