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?
For the integrand $\frac{1}{x^3+1}$, use partial fractions $\frac{2-x}{3 (x^2-x+1)}+\frac{1}{3 (x+1)}$. Integrate the sum term by term and factor out constants:
$$
I=\int \frac{1}{x^3+1} \operatorname{d}x =\frac{1}{3} \int \frac{2-x}{ x^2-x+1} \operatorname{d}x+\frac{1}{3} \int \frac{1}{x+1}\operatorname{d}x
$$
Rewrite the integrand $\frac{2-x}{ x^2-x+1}$ as $\frac{3}{2( x^2-x+1)}-\frac{2x-1}{2( x^2-x+1)}$
$$I= \frac{1}{3} \int \frac{3}{2( x^2-x+1)}\operatorname{d}x-\frac{1}{3}\int\frac{2x-1}{2( x^2-x+1)} \operatorname{d}x+\frac{1}{3} \int \frac{1}{x+1}\operatorname{d}x=I_1+I_2+I_3$$
Integrate the sum term by term and factor out constants.
For the integral $I_1=\frac{1}{3} \int \frac{3}{2( x^2-x+1)}\operatorname{d}x$, substitute $u = x^2-x+1$ and $\operatorname{d}u = (2 x-1)\operatorname{d}x$
$$I_1 = -\frac{1}{6} \int \frac{1}{u} \operatorname{d}u=-\frac{1}{6}\ln u = -\frac{1}{6}\ln(x^2-x+1)$$
For the integral $I_2=\frac{1}{3}\int\frac{2x-1}{2( x^2-x+1)} \operatorname{d}x$, complete the square
$$
I_2 = \frac{1}{2} \int \frac{1}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}
$$
and substitute $s = x-1/2$ and $ \operatorname{d}s = \operatorname{d}x$
$$I_2= \frac{1}{2} \int\frac{1}{s^2+3/4} \operatorname{d}s=\frac{2}{3} \int \frac{1}{\frac{4s^2}{3}+1} \operatorname{d}s$$
Substitute $p = 2 s/\sqrt 3$ and $\operatorname{d}p = \frac{2}{\sqrt 3} \operatorname{d}s$ so that
$$I_2= \frac{1}{\sqrt 3} \int\frac{1}{p^2+1} dp=\frac{1}{\sqrt 3}\arctan p=\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right)$$
For the integral $I_3=\frac{1}{3} \int \frac{1}{x+1}\operatorname{d}x$, substitute $w = x+1$ and $dw = dx$:
$$I_3 = \frac{1}{3} \int \frac{1}{w}\operatorname{d}w=\frac{1}{3}\ln w=\frac{1}{3}\ln (x+1)$$
Finally
$$
I(x)= -\frac{1}{6}\ln(x^2-x+1)+\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right) +\frac{1}{3}\ln (x+1)+\textrm{constant}=\frac{1}{3}\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)+\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right) +\textrm{constant}
$$
taking into account that for the logarithms: $$
-\frac{1}{6}\ln(x^2-x+1)=\frac{1}{3}\frac{1}{2}\ln\frac{1}{(x^2-x+1)}=\frac{1}{3}\ln\frac{1}{(x^2-x+1)^{\frac{1}{2}}}=\frac{1}{3}\ln\left(\frac{1}{\sqrt{x^2-x+1}}\right)
$$
so that
$$-\frac{1}{6}\ln(x^2-x+1)+\frac{1}{3}\ln (x+1)=\frac{1}{3}\ln\left(\frac{1}{\sqrt{x^2-x+1}}\right)+\frac{1}{3}\ln (x+1)=\frac{1}{3}\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)$$
The integral is
$$
\int_0^{+\infty}\frac{1}{x^3+1}\operatorname{d}x=\left[\lim_{x\to\infty}I(x)\right]-I(0)=
\frac{2\pi}{3\sqrt 3}
$$
Infact
$$
\lim_{x\to\infty}I(x)=\lim_{x\to\infty}\frac{1}{3}\ln\left(\frac{x+1}{\sqrt{x^2-x+1}}\right)+\lim_{x\to\infty}\frac{1}{\sqrt 3}\arctan \left(\frac{2x-1}{\sqrt 3} \right)
$$
and the first limit is $0$ because for $x\rightarrow\infty$ the ratio $\frac{x+1}{\sqrt{x^2-x+1}}\sim 1$ and then $\ln(1)=0$; the second limit is $\frac{1}{\sqrt 3}\frac{\pi}{2}$ because for $x\rightarrow\infty$ the function $\arctan \left(\frac{2x-1}{\sqrt 3} \right)\sim\arctan(x)$ and goes to $\pi/2$.
For $I(0)$ you have $$ I(0)=\frac{1}{3}\ln\left(\frac{0+1}{\sqrt{0^2-0+1}}\right)-\frac{1}{\sqrt 3}\arctan \left(\frac{2\cdot 0-1}{\sqrt 3} \right) =\frac{1}{3}\ln(1)-\frac{1}{\sqrt 3}\arctan \left(\frac{2\cdot 0-1}{\sqrt 3} \right)=-\frac{1}{\sqrt 3}\arctan \left(\frac{-1}{\sqrt 3} \right) $$ observing that $\arctan(1/x)+\arctan(x)=\pi/2$ and that $\arctan(-x)=-\arctan(x)$ and that $\arctan(\sqrt 3)=\pi/3$ one has $$ I(0)=-\frac{1}{\sqrt 3}\arctan \left(\frac{-1}{\sqrt 3} \right)=\frac{1}{\sqrt 3}\left(\frac{\pi}{2}-\frac{\pi}{3}\right) $$ Finally putting all together $$ \int_0^{+\infty}\frac{1}{x^3+1}\operatorname{d}x=\left[\lim_{x\to\infty}I(x)\right]-I(0)=\frac{\pi}{2\sqrt 3}+\frac{1}{\sqrt 3}\left(\frac{\pi}{2}-\frac{\pi}{3}\right)=\frac{2\pi}{3\sqrt 3} $$
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{\dd x \over 1 + x^{3}}:\ {\large ?}}$
\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}
$$\color{#00f}{\large% \int_{0}^{\infty}{\dd x \over 1 + x^{3}} = {2\root{3} \over 9}\,\pi} \approx 1.2092 $$
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 $$ \int_0^\infty\frac{1}{x^3+1}\,dx=\lim_{r\to\infty}F(r)=\frac{\pi}{2\sqrt{3}}+\frac{\pi}{6\sqrt{3}}=\frac{2\pi}{3\sqrt{3}}. $$
