<s>[INST] How to evaluate $\int_0^1\frac{\tanh ^{-1}(x)\log(x)}{(1-x) x (x+1)} \operatorname d \!x$? How to evaluate the following integral
$$\int_0^1\frac{\tanh ^{-1}(x)\log(x)}{(1-x) x (x+1)} \operatorname d \!x $$
The numerical result is $= -1.38104$ and when I look at it, I have no idea how to work on it. Could you provide me hits to evaluate the integral above?
Thank you.
 [/INST] Your integral can be rewritten as
\begin{align}
\int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x(1-x)(1+x)}{\rm d}x
=\int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x}{\rm d}x+\int^1_0\frac{x{\rm artanh}\ {x}\ln{x}}{1-x^2}{\rm d}x
\end{align}
The first integral is
\begin{align}
\int^1_0\frac{{\rm artanh}\ {x}\ln{x}}{x}{\rm d}x
=&\chi_2(x)\ln{x}\Bigg{|}^1_0-\int^1_0\frac{\chi_2(x)}{x}{\rm d}x\\
=&-\chi_3(1)=-\frac{7}{8}\zeta(3)
\end{align}
The second integral is
\begin{align}
\int^1_0\frac{x{\rm artanh}\ {x}\ln{x}}{1-x^2}{\rm d}x
=&\sum^\infty_{n=0}\sum^n_{k=0}\frac{1}{2k+1}\int^1_0 x^{2n+2}\ln{x}\ {\rm d}x\\
=&\sum^\infty_{n=0}\frac{\frac{1}{2}H_n-H_{2n+1}}{(2n+3)^2}\\
=&\frac{1}{2}\sum^\infty_{n=1}\frac{H_n}{(2n+1)^2}-\sum^\infty_{n=1}\frac{H_{2n}}{(2n+1)^2}
\end{align}
For the first sum, consider $\displaystyle f(z)=\frac{\left(\gamma+\psi_0(-z)\right)^2}{(2z+1)^2}$. At the positive integers,
\begin{align}
\sum^\infty_{n=1}{\rm Res}(f,n)
=&\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{1}{(2z+1)^2(z-n)^2}+\frac{2H_n}{(2z+1)^2(z-n)}\right]\\
=&2\sum^\infty_{n=1}\frac{H_n}{(2n+1)^2}-\frac{7}{2}\zeta(3)+4
\end{align}
At $z=0$,
\begin{align}
{\rm Res}(f,0)=\operatorname*{Res}_{z=0}\frac{1}{z^2(2z+1)^2}=-4
\end{align}
At $z=-\frac{1}{2}$,
\begin{align}
{\rm Res}\left(f,-\tfrac12\right)=\frac{1}{4}\frac{{\rm d}}{{\rm d}z}(\gamma+\psi_0(-z))^2\Bigg{|}_{z=-\frac{1}{2}}=\frac{\pi^2}{2}\ln{2}
\end{align}
Since the sum of residues is zero,
$$\sum^\infty_{n=1}\frac{H_n}{(2n+1)^2}=\frac{7}{4}\zeta(3)-\frac{\pi^2}{4}\ln{2}$$
For the second sum, consider $\displaystyle f(z)=\frac{\pi\cot(\pi z)(\gamma+\psi_0(-2z))}{(2z+1)^2}$. Note that the poles at the half integers are cancelled by the zeroes of $\pi\cot(\pi z)$. At the positive integers,
\begin{align}
\sum^\infty_{n=1}{\rm Res}(f,n)
=&\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{1}{2(2z+1)^2(z-n)^2}+\frac{H_{2n}}{(2z+1)^2(z-n)}\right]\\
=&\sum^\infty_{n=1}\frac{H_{2n}}{(2n+1)^2}-\frac{7}{4}\zeta(3)+2
\end{align}
At the negative integers,
\begin{align}
\sum^\infty_{n=1}{\rm Res}(f,-n)
=&\sum^\infty_{n=1}\frac{H_{2n+1}}{(2n+1)^2}+1\\
=&\sum^\infty_{n=1}\frac{H_{2n}}{(2n+1)^2}+\frac{7}{8}\zeta(3)
\end{align}
At $z=0$,
\begin{align}
{\rm Res}(f,0)=-2
\end{align}
Since the sum of residues is zero,
$$\sum^\infty_{n=1}\frac{H_{2n}}{(2n+1)^2}=\frac{7}{16}\zeta(3)$$
Hence, if I had not made any mistakes, the original integral is simply
$$\int^1_0\frac{\operatorname{artanh}{x}\ln{x}}{x(1-x)(1+x)}{\rm d}x=-\frac{7}{16}\zeta(3)-\frac{\pi^2}{8}\ln{2}$$
</s>