TITLE: How can I prove $\int_{0}^{1} \frac {x-1}{\log(x) (1+x^3)}dx=\frac {\log3}{2}$
QUESTION [11 upvotes]: Question:- Prove that $$\int_0^1 \frac {x-1}{\log(x) (1+x^3)} \, dx = \frac {\log(3)}{2}$$
I saw this problem as an comment on a youtube video few hours ago but I don't know how to prove this one as integration by parts doesn't works here. Also I wasn't able to figure out any proper subsitution that would simplify the integral.
Can someone suggests me some hints?

REPLY [5 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}{x - 1 \over \ln\pars{x}\pars{1 + x^{3}}}
\,\dd x} =
\int_{0}^{1}{1 \over 1 + x^{3}}\
\overbrace{\int_{0}^{1}x^{t}\,\dd t}^{\ds{x - 1 \over \ln\pars{x}}}\
\dd x
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1}{x^{t} - x^{t + 3} \over 1 - x^{6}}\,\dd x\,\dd t =
{1 \over 6}
\int_{0}^{1}\int_{0}^{1}{x^{t/6 - 5/6} - x^{t/6 - 1/3} \over 1 - x}
\,\dd x\,\dd t
\\[5mm] = &\
{1 \over 6}\int_{0}^{1}\pars{\int_{0}^{1}{1 - x^{t/6 - 1/3} \over 1 - x}
\,\dd x -
\int_{0}^{1}{1 - x^{t/6 - 5/6} \over 1 - x}
\,\dd x}\,\dd t
\\[5mm] = &\
{1 \over 6}\int_{0}^{1}\bracks{\Psi\pars{{t \over 6} + {2 \over 3}} -
\Psi\pars{{t \over 6} + {1 \over 6}}}\,\dd t =
\left. \ln\pars{\Gamma\pars{t/6 + 2/3} \over
\Gamma\pars{t/6 + 1/6}}\right\vert_{\ 0}^{\ 1}\label{1}\tag{1}
\\[5mm] = &\
\ln\pars{{\Gamma\pars{5/6} \over
\Gamma\pars{1/3}}\,{\Gamma\pars{1/6} \over
\Gamma\pars{2/3}}} =
\ln\pars{\sin\pars{\pi/3} \over \sin\pars{\pi/6}} =
\ln\pars{\root{3}/2 \over 1/2}\label{2}\tag{2}
\\[5mm] = & \bbx{\large {\ln\pars{3} \over 2}} \approx 0.5493 \\ &
\end{align}

(\ref{1}): See Digamma $\ds{\Psi}$ Identity $\ds{\bf\color{black}{6.3.22}}$.
(\ref{2}): Euler Reflection Formula $\ds{\bf\color{black}
{6.1.17}}$.
Note the Digamma $\ds{\Psi}$ Function Definition in terms of the Gamma Function $\ds{\Gamma}$:
$$
\Psi\pars{z} = \totald{\ln\pars{\Gamma\pars{z}}}{z}
$$
which was used in (\ref{1}).