How can I prove $\int_{0}^{1} \frac {x-1}{\log(x) (1+x^3)}dx=\frac {\log3}{2}$ 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?
 A: Alternatively to Mark Viola's approach, use the geometric series to see
$$\small\int_0^1\frac{x-1}{x^3+1}\frac{{\rm d}x}{\log x}=\sum_{n\ge0}(-1)^n\int_0^1\frac{x^{3n+1}-x^{3n}}{\log x}\,{\rm d}x=\sum_{n\ge0}(-1)^{n+1}\int_0^\infty\frac{e^{-(3n+2)x}-e^{-(3n+1)x}}x\,{\rm d}x$$
The latter is a Frullani integral and evaluates as
$$\int_0^\infty\frac{e^{-(3n+2)x}-e^{-(3n+1)x}}x\,{\rm d}x=-\log\left(\frac{3n+2}{3n+1}\right)$$
and thus arriving at
$$\int_0^1\frac{x-1}{x^3+1}\frac{{\rm d}x}{\log x}=\sum_{n\ge0}(-1)^n\log\left(\frac{3n+2}{3n+1}\right)$$
aswell.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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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}).
A: Note that $\int_0^1 x^s\,ds=\frac{x-1}{\log(x)}$.  Then, we have
$$\int_0^1\frac{x-1}{\log(x)(x^3+1)}\,dx=\int_0^1\int_0^1 \frac{x^s}{(x^3+1)}\,ds\,dx$$
Now we apply Fubini's Theorem to interchange the order of integration to reveal
$$\int_0^1\frac{x-1}{\log(x)(x^3+1)}\,dx=\int_0^1\int_0^1 \frac{x^s}{(x^3+1)}\,dx\,ds$$
Next, we expand the denominator in a geometric series to find that
$$\begin{align}
\int_0^1\frac{x-1}{\log(x)(x^3+1)}\,dx&=\sum_{n=0}^\infty (-1)^n\int_0^1\int_0^1 x^{s+3n}\,dx\,ds\\\\
&=\sum_{n=0}^\infty (-1)^n \log\left(\frac{3n+2}{3n+1}\right)
\end{align}$$
Can you finish now?

BONUS:
To evaluate the final series we appeal to the digamma function, its relationship with the Gamma function, and Euler's reflection formula.  Proceeding, we write
$$\begin{align}
\sum_{n=0}^\infty (-1)^n\log\left(\frac{3n+2}{3n+1}\right)&=\int_0^1 \sum_{n=0}^\infty (-1)^n \frac1{s+3n+1}\,ds\\\\
&=\int_0^1 \sum_{n=0}^\infty\left(\frac1{6n+s+1}-\frac1{6n+s+4}\right)\,ds\\\\
&=\frac16\int_0^1\left(\psi((s+4)/6)-\psi((s+1)/6)\right)\,ds\\\\
&=\log\left(\frac{\Gamma(5/6)\Gamma(1/6)}{\Gamma(2/3)\Gamma(1/3)}\right)\\\\
&=\log\left(\frac{\sin(2\pi/3)}{\sin(5\pi/6)}\right)\\\\
&=\log(\sqrt 3)
\end{align}$$
as expected!
A: Note
$$I=\int_{0}^{1} \frac {x-1}{\ln x (1+x^3)}dx
\overset{x\to\frac1x}=
\frac12\int_{0}^{\infty} \frac {x-1}{\ln x (1+x^3)}dx$$
Let $J(a) = \int_{0}^{\infty} \frac {x^a-1}{\ln x (1+x^3)}dx$. Then
$J’(a) = \int_{0}^{\infty} \frac {x^a}{1+x^3}dx=\frac\pi3\csc\frac{\pi(a+1)}3
$.
Thus,
$$I=\frac12 J(1) =\frac12\int_0^1J’(a)da=\frac\pi6\int_0^1\csc\frac{\pi(a+1)}3da=\frac{\ln3}2
$$
