Prove $\int_0^{\infty} \frac{\ln^2(x^3+1)}{x^3+1} dx = \frac{\sqrt{3} \pi}{18} \left(9\ln^2(3)+4\psi ^{\prime} \left(\frac{2}{3}\right)\right)-\ldots$ Prove $$\int_0^{\infty} \frac{\ln^2{(x^3+1)}}{x^3+1} \; \mathrm{d}x = \frac{\sqrt{3} \pi}{18} \left(9\ln^2{(3)}+4\psi ^{\prime} \left(\frac{2}{3}\right)\right)-\frac{\pi^3 \sqrt{3}}{54}-\frac{\pi}{3}\ln{(3)}$$
I tried feynman method with $ I(a)=\int_0^{\infty} \frac{\ln^2\left(ax^3+1\right)}{x^3+1} \; \mathrm{d}x$ but this got ugly because we need to differentiate wrt a twice I think.  I also try to factor $x^3+1$ and either partial fraction decomposition denominator or log property for numerator but these did not work.  I am not sure what to do now. Im not very good with contour integration so can people who respond try to use real methods?? Any help is appreciated
 A: Consider the parameterized integral $I(a)$ where the integral in question is equal to $I''(1)$:
$$I(a)=\int_0^{\infty} \frac{1}{\left(x^3+1\right)^a} \; \mathrm{d}x$$
First, let $x^3 \to x$:
\begin{align}
I(a) &= \frac{1}{3} \int_0^{\infty} \frac{t^{-\frac{2}{3}}}{(t+1)^a} \notag \\
 & \; \; = \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)}{\Gamma\left(a\right)} \notag 
\end{align}
Where we used the definition of the beta function.
Now we will find $I''(1)$:
\begin{align}
I'(a) &= \frac{d}{da} \left( \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)}{\Gamma\left(a\right)} \right) \notag \\
 &= \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)\left(\psi\left(a-\frac{1}{3}\right)-\psi\left(a\right)\right)}{\Gamma\left(a\right)} \notag \\
I''(a) &= \frac{d}{da} \left( \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)\left(\psi\left(a-\frac{1}{3}\right)-\psi\left(a\right)\right)}{\Gamma\left(a\right)} \right) \notag \\ 
&= \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(a-\frac{1}{3}\right)\left(\psi\left(a-\frac{1}{3}\right)^2-2\psi\left(a\right)\psi\left(a-\frac{1}{3}\right)+\psi\left(a\right)^2+\psi^{\prime}\left(a-\frac{1}{3}\right)-\psi^{\prime}\left(a\right)\right)}{\Gamma\left(a\right)} \notag \\
I''(1) &= \frac{\Gamma\left(\frac{4}{3}\right)\Gamma\left(\frac{2}{3}\right)\left(\psi\left(\frac{2}{3}\right)^2-2\psi\left(1\right)\psi\left(\frac{2}{3}\right)+\psi\left(1\right)^2+\psi^{\prime}\left(\frac{2}{3}\right)-\psi^{\prime}\left(1\right)\right)}{\Gamma\left(1\right)} \notag \\
 &= \frac{2\pi}{3\sqrt{3}}\left(-\frac{\pi^2}{12}+\frac{9\ln^2{(3)}}{4}-\frac{\pi \sqrt{3} }{2} \ln{(3)} + \psi^{\prime} \left(\frac{2}{3}\right)\right) \notag \\
 &= \boxed{\frac{\sqrt{3} \pi}{18} \left(9\ln^2{(3)}+4\psi ^{\prime} \left(\frac{2}{3}\right)\right)-\frac{\pi^3 \sqrt{3}}{54}-\color{red}{\frac{\pi^2}{3}\ln{(3)}}} \notag
\end{align}
It appears that the result of the integral you provided is slightly off (see the red term).  Wolfram Alpha agrees with the answer I provided as well.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x^{3} + 1} \over x^{3} + 1}\,\dd x}
\,\,\,\stackrel{\large x^{3}\ \mapsto\ x}{=}\,\,\,
\int_{0}^{\infty}{\ln^{2}\pars{x + 1} \over x + 1}\,
{1 \over 3}\,x^{-2/3}\,\dd x
\\[5mm] = &\
\left.{1 \over 3}\,\partiald[2]{}{\nu}\int_{0}^{\infty}x^{\color{red}{1/3} - 1}\,\pars{x + 1}^{\nu - 1}\,\dd x\,\right\vert_{\ \nu\ =\ 0}\label{1}\tag{1}
\end{align}
\begin{align}
\mbox{However,}\quad
\pars{x + 1}^{\nu - 1} & = \sum_{k = 0}^{\infty}
{\nu - 1 \choose k}x^{k} =
\sum_{k = 0}^{\infty}\bracks{{k - \nu \choose k}\pars{-1}^{k}}x^{k}
\\[5mm] &=
\sum_{k = 0}^{\infty}
\color{red}{\Gamma\pars{k - \nu + 1} \over \Gamma\pars{1 - \nu}}\,
{\pars{-x}^{k} \over k!}
\end{align}
With the Ramanujan Master Theorem, (\ref{1}) becomes:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x^{3} + 1} \over x^{3} + 1}\,\dd x} =
\left.{1 \over 3}\,\partiald[2]{}{\nu}
\bracks{\Gamma\pars{1 \over 3}{\Gamma\pars{-1/3 - \nu + 1} \over \Gamma\pars{1 - \nu}}}
\,\right\vert_{\ \nu\ =\ 0}
\\[5mm] = &\
\bbx{{2 \over 9}\,\root{3}\pi\,\Psi\, '\pars{2 \over 3} +
{1 \over 2}\,\root{3}\pi\ln^{2}\pars{3} -
{1 \over 3}\pi^{2}\ln\pars{3} - {1 \over 54}\,\root{3}\pi^{3}}
\\[5mm] \approx &\ 2.3798
\end{align}
