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![\[\int_0^1{\ln ^3\left( 1-x \right) \ln ^2\left( 1+x \right) \text{d}x}=2\underset{I}{\underbrace{\int_0^1{\frac{\ln ^2\left( \dfrac{2}{1+x} \right) \ln ^3\left( \dfrac{2x}{1+x} \right)}{\left( 1+x \right) ^2}\text{d}x}}}\]](http://latex.artofproblemsolving.com/a/f/a/afa4f7eada49a43e847ba9d591deb49032e9c900.png)
\begin{align*} &\ln ^2\left( \frac{2}{1+x} \right) \ln ^3\left( \frac{2x}{1+x} \right) =-\ln ^2\text{2}\ln ^3\left( 1+x \right) +\text{2}\ln\text{2}\ln ^4\left( 1+x \right) -\ln ^5\left( 1+x \right) \\ &+\text{3}\ln ^2\text{2}\ln ^2\left( 1+x \right) \ln \left( 2x \right) -\text{6}\ln\text{2}\ln ^3\left( 1+x \right) \ln \left( 2x \right) +\text{3}\ln ^4\left( 1+x \right) \ln \left( 2x \right) \\ &-\text{3}\ln ^2\text{2}\ln \left( 1+x \right) \ln ^2\left( 2x \right) +\text{6}\ln\text{2}\ln ^2\left( 1+x \right) \ln ^2\left( 2x \right) -\text{3}\ln ^3\left( 1+x \right) \ln ^2\left( 2x \right) \\ &+\ln ^2\text{2}\ln ^3\left( 2x \right) -\text{2}\ln\text{2}\ln \left( 1+x \right) \ln ^3\left( 2x \right) +\ln ^2\left( 1+x \right) \ln ^3\left( 2x \right) \end{align*}
![\[I_1=-\ln ^22\int_0^1{\frac{\ln ^3\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=\frac{\ln ^52}{2}+\frac{\text{3}\ln ^42}{2}+\text{3}\ln ^32-\text{3}\ln ^22\]](http://latex.artofproblemsolving.com/e/5/b/e5b9fe12236c0975bca1f8f990c7a5e2bd163d45.png)
![\[I_2=\text{2}\ln 2\int_0^1{\frac{\ln ^4\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=-\ln ^52-\text{4}\ln ^42-\text{12}\ln ^32-\text{24}\ln ^22+\text{24}\ln 2\]](http://latex.artofproblemsolving.com/0/e/8/0e89d68abc71fabfe11d32310826b29783009b4c.png)
![\[I_3=-\int_0^1{\frac{\ln ^5\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=-60+\frac{\ln ^52}{2}+\frac{\text{5}\ln ^42}{2}+\text{10}\ln ^32+\text{30}\ln ^22+\text{60}\ln 2\]](http://latex.artofproblemsolving.com/c/8/3/c83cafe6abaf11840cd25b6fd2c366cd0a92c8d7.png)
\begin{align*} I_4&=\text{3}\ln ^22\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln \left( 2x \right)}{\left( 1+x \right) ^2}}\text{d}x\\&=\text{3}\ln ^22\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x+\text{3}\ln ^32\int_0^1{\frac{\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x \end{align*}
![\[I_{41}=\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x=\frac{1}{4}\zeta \left( 3 \right) +\frac{\pi ^2}{6}-\frac{1}{3}\ln ^32-\ln ^22-\text{2}\ln 2\]](http://latex.artofproblemsolving.com/0/7/4/074a6ac27ad46f57a9c2b6c4fc2effa355e5fcef.png)
![\[I_{42}=I_{82}=\int_0^1{\frac{\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=1-\frac{1}{2}\ln ^22-\ln 2\]](http://latex.artofproblemsolving.com/1/1/1/111dc96f9ebebb3efebc8963f0ac74530be4c547.png)
\begin{align*} I_5&=-\text{6}\ln 2\int_0^1{\frac{\ln ^3\left( 1+x \right) \ln \left( 2x \right)}{\left( 1+x \right) ^2}\text{d}x}\\ &=-\text{6}\ln 2\int_0^1{\frac{\ln ^3\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}\text{d}x}-\text{6}\ln ^22\int_0^1{\frac{\ln ^3\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x} \end{align*}
\begin{align*} I_{51}&=\int_0^1{\frac{\ln ^3\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}\text{d}x}=\int_0^1{\frac{\ln ^3\left( 1+x \right)}{x\left( 1+x \right)}\text{d}x}+3\int_0^1{\frac{\ln x\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x\\ &=\int_0^1{\frac{\ln ^3\left( 1+x \right)}{x}\text{d}x}-\int_0^1{\frac{\ln ^3\left( 1+x \right)}{1+x}\text{d}x}+3\int_0^1{\frac{\ln x\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x\\ &=\left[ -\text{6Li}_4\left( \frac{1}{2} \right) -\frac{21}{4}\ln 2\zeta \left( 3 \right) +\frac{1}{15}\pi ^4-\frac{1}{4}\ln ^42+\frac{1}{4}\pi ^2\ln ^22 \right] -\frac{\ln ^42}{4} \\&+3\left[ \frac{1}{4}\zeta \left( 3 \right) +\frac{\pi ^2}{6}-\frac{1}{3}\ln ^32-\ln ^22-\text{2}\ln 2 \right] \\&=-\text{6Li}_4\left( \frac{1}{2} \right) -\frac{21}{4}\ln 2\zeta \left( 3 \right) +\frac{1}{15}\pi ^4-\frac{1}{2}\ln ^42-\ln ^32+\frac{1}{4}\pi ^2\ln ^22\\ &+\frac{3}{4}\zeta \left( 3 \right) +\frac{\pi ^2}{2}-\text{3}\ln ^22-\text{6}\ln 2 \end{align*}
![\[I_{52}=\int_0^1{\frac{\ln ^3\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=3-\frac{1}{2}\ln ^32-\frac{3}{2}\ln ^22-\text{3}\ln 2\]](http://latex.artofproblemsolving.com/6/8/a/68a3ded6b6c0ac9c5f8f123470a21f409f63d9c3.png)
![\[I_6=3\int_0^1{\frac{\ln ^4\left( 1+x \right) \ln \left( 2x \right)}{\left( 1+x \right) ^2}\text{d}x}=3\int_0^1{\frac{\ln ^4\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}\text{d}x}+\text{3}\ln 2\int_0^1{\frac{\ln ^4\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}\]](http://latex.artofproblemsolving.com/b/5/c/b5cd5c75ec946855a9d0039eacae8fc03f15b2e8.png)
\begin{align*} I_{61}&=\int_0^1{\frac{\ln ^4\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}\text{d}x}=\int_0^1{\frac{\ln ^4\left( 1+x \right)}{x\left( 1+x \right)}\text{d}x}+4\int_0^1{\frac{\ln x\ln ^3\left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x\\ &=\int_0^1{\frac{\ln ^4\left( 1+x \right)}{x}\text{d}x}-\int_0^1{\frac{\ln ^4\left( 1+x \right)}{1+x}\text{d}x}+4\int_0^1{\frac{\ln x\ln ^3\left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x \\&=\left[ -\text{24Li}_5\left( \frac{1}{2} \right) -\text{24}\ln\text{2Li}_4\left( \frac{1}{2} \right) +24\zeta \left( 5 \right) -\frac{21}{2}\ln ^22\zeta \left( 3 \right) -\frac{4}{5}\ln ^52+\frac{2}{3}\pi ^2\ln ^32 \right] -\frac{1}{5}\ln ^52 \\&+4\left[ -\text{6Li}_4\left( \frac{1}{2} \right) -\frac{21}{4}\ln 2\zeta \left( 3 \right) +\frac{1}{15}\pi ^4-\frac{1}{2}\ln ^42-\ln ^32+\frac{1}{4}\pi ^2\ln ^22+\frac{3}{4}\zeta \left( 3 \right) +\frac{\pi ^2}{2}-\text{3}\ln ^22-\text{6}\ln 2 \right] \\&=-\text{24Li}_5\left( \frac{1}{2} \right) -\text{24Li}_4\left( \frac{1}{2} \right) -\text{24}\ln\text{2Li}_4\left( \frac{1}{2} \right) +24\zeta \left( 5 \right) -\frac{21}{2}\ln ^22\zeta \left( 3 \right) -\text{21}\ln 2\zeta \left( 3 \right) -\ln ^52+\frac{4}{15}\pi ^4 \\&-\text{2}\ln ^42-\text{4}\ln ^32+\frac{2}{3}\pi ^2\ln ^32+\pi ^2\ln ^22+3\zeta \left( 3 \right) +2\pi ^2-\text{12}\ln ^22-\text{24}\ln 2 \end{align*}
![\[I_{62}=\int_0^1{\frac{\ln ^4\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=12-\frac{1}{2}\ln ^42-\text{2}\ln ^32-\text{6}\ln ^22-\text{12}\ln 2\]](http://latex.artofproblemsolving.com/5/b/1/5b1af70657bcd6de3da6ebaa2f465e219c6ceda0.png)
\begin{align*} I_7&=-\text{3}\ln ^22\int_0^1{\frac{\ln \left( 1+x \right) \ln ^2\left( 2x \right)}{\left( 1+x \right) ^2}}\text{d}x=-\text{3}\ln ^22\int_0^1{\frac{\ln \left( 1+x \right) \ln ^2x}{\left( 1+x \right) ^2}}\text{d}x\\ &-\text{3}\ln ^42\int_0^1{\frac{\ln \left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x-\text{6}\ln ^32\int_0^1{\frac{\ln \left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x \end{align*}
\begin{align*} I_{71}&=\int_0^1{\frac{\ln \left( 1+x \right) \ln ^2x}{\left( 1+x \right) ^2}}\text{d}x=\int_0^1{\frac{\ln ^2x}{\left( 1+x \right) ^2}}\text{d}x+2\int_0^1{\frac{\ln \left( 1+x \right) \ln x}{x\left( 1+x \right)}\text{d}x}\\ &=\int_0^1{\frac{\ln ^2x}{\left( 1+x \right) ^2}}\text{d}x+2\int_0^1{\frac{\ln \left( 1+x \right) \ln x}{x}\text{d}x}-2\int_0^1{\frac{\ln \left( 1+x \right) \ln x}{1+x}\text{d}x} \\&=\frac{1}{6}\pi ^2+2\left[ -\frac{3}{4}\zeta \left( 3 \right) \right] -2\left[ -\frac{1}{8}\zeta \left( 3 \right) \right] =-\frac{5}{4}\zeta \left( 3 \right) +\frac{1}{6}\pi ^2 \end{align*}
![\[I_{72}=\int_0^1{\frac{\ln \left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x=\frac{1}{2}-\frac{1}{2}\ln 2\]](http://latex.artofproblemsolving.com/8/3/c/83c930ccef7762f8c58c31d9808aa1dd88d13813.png)
![\[I_{73}=\int_0^1{\frac{\ln \left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x=\frac{1}{12}\pi ^2-\frac{1}{2}\ln ^22-\ln 2\]](http://latex.artofproblemsolving.com/a/f/d/afd497b41d2bcffa66707ac1d1440a91b7d2c51e.png)
\begin{align*} I_8&=\text{6}\ln 2\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln ^2\left( 2x \right)}{\left( 1+x \right) ^2}\text{d}x}=\text{6}\ln 2\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln ^2x}{\left( 1+x \right) ^2}\text{d}x}\\&+\text{6}\ln ^32\int_0^1{\frac{\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}+\text{12}\ln ^22\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}\text{d}x} \end{align*}
\begin{align*} I_{81}&=\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln ^2x}{\left( 1+x \right) ^2}\text{d}x}=2\int_0^1{\frac{\ln \left( 1+x \right) \ln ^2x}{\left( 1+x \right) ^2}\text{d}x}+2\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{x\left( 1+x \right)}\text{d}x}\\ &=2\int_0^1{\frac{\ln ^2x}{\left( 1+x \right) ^2}\text{d}x}+4\int_0^1{\frac{\ln \left( 1+x \right) \ln x}{x\left( 1+x \right)}}\text{d}x+2\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{x}\text{d}x}-2\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{1+x}\text{d}x} \\&=2\int_0^1{\frac{\ln ^2x}{\left( 1+x \right) ^2}\text{d}x}+4\int_0^1{\frac{\ln \left( 1+x \right) \ln x}{x}}\text{d}x-4\int_0^1{\frac{\ln \left( 1+x \right) \ln x}{1+x}}\text{d}x+2\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{x}\text{d}x}-2\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{1+x}\text{d}x} \\&=2\cdot \frac{1}{6}\pi ^2+4\left[ -\frac{3}{4}\zeta \left( 3 \right) \right] -4\left[ -\frac{1}{8}\zeta \left( 3 \right) \right] +2\left[ -\text{4Li}_4\left( \frac{1}{2} \right) -\frac{7}{2}\ln 2\zeta \left( 3 \right) +\frac{1}{24}\pi ^4-\frac{1}{6}\ln ^42+\frac{1}{6}\pi ^2\ln ^22 \right] \\&-2\left[ \text{2Li}_4\left( \frac{1}{2} \right) +\frac{7}{4}\ln 2\zeta \left( 3 \right) -\frac{1}{45}\pi ^4+\frac{1}{12}\ln ^42-\frac{1}{12}\pi ^2\ln ^22 \right] \\&=-\text{12Li}_4\left( \frac{1}{2} \right) -\frac{21}{2}\ln 2\zeta \left( 3 \right) -\frac{5}{2}\zeta \left( 3 \right) +\frac{23}{180}\pi ^4-\frac{1}{2}\ln ^42+\frac{1}{2}\pi ^2\ln ^22+\frac{1}{3}\pi ^2 \end{align*}
![\[I_{82}=\int_0^1{\frac{\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=1-\frac{1}{2}\ln ^22-\ln 2\]](http://latex.artofproblemsolving.com/d/d/2/dd2c1647106ec93cff1ebfdcdf2a212adc9bd830.png)
\begin{align*} I_{83}&=\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}\text{d}x}=\int_0^1{\frac{\ln ^2\left( 1+x \right)}{x\left( 1+x \right)}\text{d}x}+2\int_0^1{\frac{\ln x\ln \left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x\\ &=\int_0^1{\frac{\ln ^2\left( 1+x \right)}{x}\text{d}x}-\int_0^1{\frac{\ln ^2\left( 1+x \right)}{1+x}\text{d}x}+2\int_0^1{\frac{\ln x}{\left( 1+x \right) ^2}\text{d}x}+2\int_0^1{\frac{\ln \left( 1+x \right)}{x\left( 1+x \right)}}\text{d}x \\&=\int_0^1{\frac{\ln ^2\left( 1+x \right)}{x}\text{d}x}-\int_0^1{\frac{\ln ^2\left( 1+x \right)}{1+x}\text{d}x}+2\int_0^1{\frac{\ln x}{\left( 1+x \right) ^2}\text{d}x}+2\int_0^1{\frac{\ln \left( 1+x \right)}{x}}\text{d}x-2\int_0^1{\frac{\ln \left( 1+x \right)}{1+x}}\text{d}x \\&=\frac{1}{4}\zeta \left( 3 \right) -\frac{\ln ^32}{3}+2\left( -\ln 2 \right) +2\cdot \frac{\pi ^2}{12}-2\cdot \frac{\ln ^22}{2}=\frac{1}{4}\zeta \left( 3 \right) +\frac{\pi ^2}{6}-\frac{1}{3}\ln ^32-\ln ^22-\text{2}\ln 2 \end{align*}
\begin{align*} I_9&=-3\int_0^1{\frac{\ln ^3\left( 1+x \right) \ln ^2\left( 2x \right)}{\left( 1+x \right) ^2}}\text{d}x=-3\int_0^1{\frac{\ln ^3\left( 1+x \right) \ln ^2x}{\left( 1+x \right) ^2}}\text{d}x\\&-\text{3}\ln ^22\int_0^1{\frac{\ln ^3\left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x-\text{3}\ln 2\int_0^1{\frac{\ln ^3\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x \end{align*}
\begin{align*} I_{91}&=\int_0^1{\frac{\ln ^3\left( 1+x \right) \ln ^2x}{\left( 1+x \right) ^2}}\text{d}x=2\int_0^1{\frac{\ln x\ln ^3\left( 1+x \right)}{x\left( 1+x \right)}\text{d}x}+3\int_0^1{\frac{\ln ^2x\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}\\ &=2\int_0^1{\frac{\ln x\ln ^3\left( 1+x \right)}{x}\text{d}x}-2\int_0^1{\frac{\ln x\ln ^3\left( 1+x \right)}{1+x}\text{d}x}+3\int_0^1{\frac{\ln ^2x\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x} \\&=2\left[ -\text{12Li}_5\left( \frac{1}{2} \right) -\text{12}\ln\text{2Li}_4\left( \frac{1}{2} \right) +\frac{1}{2}\pi ^2\zeta \left( 3 \right) +\frac{99}{16}\zeta \left( 5 \right) -\frac{2}{5}\ln ^52+\frac{1}{3}\pi ^2\ln ^32-\frac{21}{4}\ln ^22\zeta \left( 3 \right) \right] \\&-2\left[ \text{6Li}_5\left( \frac{1}{2} \right) +\text{6}\ln\text{2Li}_4\left( \frac{1}{2} \right) -6\zeta \left( 5 \right) +\frac{21}{8}\ln ^22\zeta \left( 3 \right) +\frac{1}{5}\ln ^52-\frac{1}{6}\pi ^2\ln ^32 \right] \\&+3\left[ -\text{12Li}_4\left( \frac{1}{2} \right) -\frac{21}{2}\ln 2\zeta \left( 3 \right) -\frac{5}{2}\zeta \left( 3 \right) +\frac{23}{180}\pi ^4-\frac{1}{2}\ln ^42+\frac{1}{2}\pi ^2\ln ^22+\frac{1}{3}\pi ^2 \right] \\&=-\text{36Li}_5\left( \frac{1}{2} \right) -\text{36Li}_4\left( \frac{1}{2} \right) -\text{36}\ln\text{2Li}_4\left( \frac{1}{2} \right) +\pi ^2\zeta \left( 3 \right) +\frac{195}{8}\zeta \left( 5 \right) -\frac{15}{2}\zeta \left( 3 \right) -\frac{6}{5}\ln ^52+\pi ^2\ln ^32 \\&-\frac{63}{4}\ln ^22\zeta \left( 3 \right) -\frac{63}{2}\ln 2\zeta \left( 3 \right) +\frac{23}{60}\pi ^4-\frac{3}{2}\ln ^42+\frac{3}{2}\pi ^2\ln ^22+\pi ^2 \end{align*}
![\[I_{92}=I_{52}=\int_0^1{\frac{\ln ^3\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=3-\frac{1}{2}\ln ^32-\frac{3}{2}\ln ^22-\text{3}\ln 2\]](http://latex.artofproblemsolving.com/1/3/1/131e618e818fbe6a8dbabac96e79be5063d93a84.png)
\begin{align*} I_{93}&=I_{51}=\int_0^1{\frac{\ln ^3\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x \\&=-\text{6Li}_4\left( \frac{1}{2} \right) -\frac{21}{4}\ln 2\zeta \left( 3 \right) +\frac{1}{15}\pi ^4-\frac{1}{2}\ln ^42-\ln ^32+\frac{1}{4}\pi ^2\ln ^22+\frac{3}{4}\zeta \left( 3 \right) +\frac{\pi ^2}{2}-\text{3}\ln ^22-\text{6}\ln 2 \end{align*}
![\[I_{10}=\ln ^22\int_0^1{\frac{\ln ^3\left( 2x \right)}{\left( 1+x \right) ^2}}\text{d}x=-\frac{9}{2}\ln ^22\zeta \left( 3 \right) -\frac{5}{2}\ln ^52+\frac{1}{2}\pi ^2\ln ^32\]](http://latex.artofproblemsolving.com/2/0/5/205e121986ba5a7a98eb9bbee136dac126bddc91.png)
\begin{align*} I_{11}&=-\text{2}\ln 2\int_0^1{\frac{\ln \left( 1+x \right) \ln ^3\left( 2x \right)}{\left( 1+x \right) ^2}}\text{d}x=-\text{2}\ln ^42\int_0^1{\frac{\ln \left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x-\text{6}\ln ^32\int_0^1{\frac{\ln \left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x \\&-\text{6}\ln ^22\int_0^1{\frac{\ln \left( 1+x \right) \ln ^2x}{\left( 1+x \right) ^2}}\text{d}x-\text{2}\ln 2\int_0^1{\frac{\ln \left( 1+x \right) \ln ^3x}{\left( 1+x \right) ^2}}\text{d}x \end{align*}
![\[I_{111}=I_{72}=\int_0^1{\frac{\ln \left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x=\frac{1}{2}-\frac{1}{2}\ln 2\]](http://latex.artofproblemsolving.com/9/d/5/9d5cb3824d329d13c3d3a3d026a56b579d8d37b3.png)
![\[I_{112}=I_{73}=\int_0^1{\frac{\ln \left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x=\frac{1}{12}\pi ^2-\frac{1}{2}\ln ^22-\ln 2\]](http://latex.artofproblemsolving.com/d/a/8/da8fc8322e32a756807a7ebd8eb8821540576058.png)
![\[I_{113}=I_{71}=\int_0^1{\frac{\ln \left( 1+x \right) \ln ^2x}{\left( 1+x \right) ^2}}\text{d}x=-\frac{5}{4}\zeta \left( 3 \right) +\frac{1}{6}\pi ^2\]](http://latex.artofproblemsolving.com/d/7/2/d722a99467de9db3ad1ba5d051b08d51888af6d1.png)
\begin{align*} I_{114}&=\int_0^1{\frac{\ln \left( 1+x \right) \ln ^3x}{\left( 1+x \right) ^2}}\text{d}x=\int_0^1{\frac{\ln ^3x}{\left( 1+x \right) ^2}\text{d}x}+3\int_0^1{\frac{\ln \left( 1+x \right) \ln ^2x}{x\left( 1+x \right)}}\text{d}x \\&=\int_0^1{\frac{\ln ^3x}{\left( 1+x \right) ^2}\text{d}x}+3\int_0^1{\frac{\ln \left( 1+x \right) \ln ^2x}{x}}\text{d}x-3\int_0^1{\frac{\ln \left( 1+x \right) \ln ^2x}{1+x}}\text{d}x \\&=-\frac{9}{2}\zeta \left( 3 \right) +3\cdot \frac{7\pi ^4}{360}-3\left[ \text{4Li}_4\left( \frac{1}{2} \right) +\frac{7}{2}\ln 2\zeta \left( 3 \right) -\frac{1}{24}\pi ^4+\frac{1}{6}\ln ^42-\frac{1}{6}\pi ^2\ln ^22 \right] \\&=-\text{12Li}_4\left( \frac{1}{2} \right) -\frac{21}{2}\ln 2\zeta \left( 3 \right) -\frac{9}{2}\zeta \left( 3 \right) +\frac{11}{60}\pi ^4-\frac{1}{2}\ln ^42+\frac{1}{2}\pi ^2\ln ^22 \end{align*}
\begin{align*} I_{12}&=\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln ^3\left( 2x \right)}{\left( 1+x \right) ^2}}\text{d}x=\ln ^32\int_0^1{\frac{\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}}\text{d}x+\text{3}\ln ^22\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x \\&+\text{3}\ln 2\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln ^2x}{\left( 1+x \right) ^2}}\text{d}x+\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln ^3x}{\left( 1+x \right) ^2}}\text{d}x \end{align*}
![\[I_{121}=I_{42}=I_{82}=\int_0^1{\frac{\ln ^2\left( 1+x \right)}{\left( 1+x \right) ^2}\text{d}x}=1-\frac{1}{2}\ln ^22-\ln 2\]](http://latex.artofproblemsolving.com/3/9/e/39e6a43e083d8fe70268ea7e71bd19f2b736cdbf.png)
![\[I_{122}=I_{41}=\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln x}{\left( 1+x \right) ^2}}\text{d}x=\frac{1}{4}\zeta \left( 3 \right) +\frac{\pi ^2}{6}-\frac{1}{3}\ln ^32-\ln ^22-\text{2}\ln 2\]](http://latex.artofproblemsolving.com/5/8/6/5867408d63ef6ea105e1775473e2ccc976a1d235.png)
![\[I_{123}=I_{81}=\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln ^2x}{\left( 1+x \right) ^2}\text{d}x}=-\text{12Li}_4\left( \frac{1}{2} \right) -\frac{21}{2}\ln 2\zeta \left( 3 \right) -\frac{5}{2}\zeta \left( 3 \right) +\frac{23}{180}\pi ^4-\frac{1}{2}\ln ^42+\frac{1}{2}\pi ^2\ln ^22+\frac{1}{3}\pi ^2\]](http://latex.artofproblemsolving.com/3/4/2/34211b80b53e9a3ad9090b75a687980cc6daf976.png)
\begin{align*} I_{124}&=\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln ^3x}{\left( 1+x \right) ^2}}\text{d}x=2\int_0^1{\frac{\ln \left( 1+x \right) \ln ^3x}{\left( 1+x \right) ^2}\text{d}x}+3\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln ^2x}{x\left( 1+x \right)}}\text{d}x \\&=2\int_0^1{\frac{\ln \left( 1+x \right) \ln ^3x}{\left( 1+x \right) ^2}\text{d}x}+3\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln ^2x}{x}}\text{d}x-3\int_0^1{\frac{\ln ^2\left( 1+x \right) \ln ^2x}{1+x}}\text{d}x \\&=2\left[ -\text{12Li}_4\left( \frac{1}{2} \right) -\frac{21}{2}\ln 2\zeta \left( 3 \right) -\frac{9}{2}\zeta \left( 3 \right) +\frac{11}{60}\pi ^4-\frac{1}{2}\ln ^42+\frac{1}{2}\pi ^2\ln ^22 \right] +3\left[ \frac{1}{3}\pi ^2\zeta \left( 3 \right) -\frac{29}{5}\zeta \left( 5 \right) \right] \\&-3\left[ \text{8Li}_5\left( \frac{1}{2} \right) +\text{8}\ln\text{2Li}_4\left( \frac{1}{2} \right) -\frac{33}{8}\zeta \left( 5 \right) -\frac{1}{3}\pi ^2\zeta \left( 3 \right) +\frac{7}{2}\ln ^22\zeta \left( 3 \right) +\frac{4}{15}\ln ^52-\frac{2}{9}\pi ^2\ln ^32 \right] \\&=-\text{24Li}_5\left( \frac{1}{2} \right) -\text{24Li}_4\left( \frac{1}{2} \right) -\text{24}\ln\text{2Li}_4\left( \frac{1}{2} \right) -\frac{201}{40}\zeta \left( 5 \right) -\frac{21}{2}\ln ^22\zeta \left( 3 \right) -\text{21}\ln 2\zeta \left( 3 \right) +2\pi ^2\zeta \left( 3 \right) \\&-9\zeta \left( 3 \right) +\frac{11}{30}\pi ^4-\ln ^42+\pi ^2\ln ^22-\frac{4}{5}\ln ^52+\frac{2}{3}\pi ^2\ln ^32 \end{align*}
