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{\displaystyle \int _{0}^{1}{\frac {{\text{artanh}}\,x\,\,\log x}{x\,(1-x)\,(1+x)}}\,dx=-{\frac {1}{16}}{\Big (}7\zeta (3)+2\pi ^{2}\log 2{\Big )}}

Beweis

{\displaystyle {\text{artanh}}\,x={\frac {1}{2}}\cdot \log \left({\frac {1+x}{1-x}}\right)={\frac {1}{2}}\cdot {\Big [}\log(1+x)-\log(1-x){\Big ]}}

{\displaystyle \Rightarrow \,{\text{artanh}}\,x\cdot \log x={\frac {1}{2}}\cdot \log(1+x)\log x-{\frac {1}{2}}\cdot \log(1-x)\log x}

{\displaystyle {\frac {1}{x\,(1-x)\,(1+x)}}={\frac {1}{x}}+{\frac {1}{2}}\cdot {\frac {1}{1-x}}-{\frac {1}{2}}\cdot {\frac {1}{1+x}}}


{\displaystyle {\begin{aligned}{\frac {{\text{artanh}}\,x\cdot \log x}{x\,(1-x)\,(1+x)}}=&+{\frac {1}{2}}\cdot {\frac {\log(1+x)\log x}{x}}+{\frac {1}{4}}\cdot {\frac {\log(1+x)\log x}{1-x}}-{\frac {1}{4}}\cdot {\frac {\log(1+x)\log x}{1+x}}\\\\&-{\frac {1}{2}}\cdot {\frac {\log(1-x)\log x}{x}}-{\frac {1}{4}}\cdot {\frac {\log(1-x)\log x}{1-x}}+{\frac {1}{4}}\cdot {\frac {\log(1-x)\log x}{1+x}}\end{aligned}}}



{\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {{\text{artanh}}\,x\cdot \log x}{x\,(1-x)\,(1+x)}}\,dx=&+{\frac {1}{2}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1+x)\log x}{x}}\,dx} _{=-{\frac {3}{4}}\zeta (3)}+{\frac {1}{4}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1+x)\log x}{1-x}}\,dx} _{=-{\frac {\pi ^{2}}{4}}\log 2+\zeta (3)}-{\frac {1}{4}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1+x)\log x}{1+x}}\,dx} _{-{\frac {1}{8}}\zeta (3)}\\\\&-{\frac {1}{2}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1-x)\log x}{x}}\,dx} _{=\zeta (3)}-{\frac {1}{4}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1-x)\log x}{1-x}}\,dx} _{=\zeta (3)}+{\frac {1}{4}}\cdot \underbrace {\int _{0}^{1}{\frac {\log(1-x)\log x}{1+x}}\,dx} _{=-{\frac {\pi ^{2}}{4}}\log 2+{\frac {13}{8}}\zeta (3)}\end{aligned}}}


{\displaystyle \int _{0}^{1}{\frac {{\text{artanh}}\,x\cdot \log x}{x\,(1-x)\,(1+x)}}\,dx=-{\frac {3}{8}}\zeta (3)-{\frac {\pi ^{2}}{16}}\log 2+{\frac {1}{4}}\zeta (3)+{\frac {1}{32}}\zeta (3)-{\frac {1}{2}}\zeta (3)-{\frac {1}{4}}\zeta (3)-{\frac {\pi ^{2}}{16}}\log 2+{\frac {13}{32}}\zeta (3)=-{\frac {7}{16}}\zeta (3)-{\frac {\pi ^{2}}{8}}\log 2}