Off the charts!

$\begin{matrix} \text{let }I=\int_0^1 {\dfrac {(\ln x)^3}{x^2- 1} dx} & & & \color{#D61F06} \text{put }x=e^{-u} \end{matrix} \\\Rightarrow \begin{matrix} I=\int_0^{\infty} {\dfrac {u^3e^{-u}du}{1-e^{-2u}}} & & & \color{#D61F06} \dfrac {1}{1-e^{-2u}}=\displaystyle\sum_{k=0}^\infty{e^{-2kx}} & \color{#D61F06} \text{sum geometric series} \end{matrix} \\\Rightarrow \begin{matrix} I=\int_0^{\infty} {\displaystyle\sum_{k=0}^\infty{u^3e^{-u}e^{-2ku}}du}= \displaystyle\sum_{k=0}^\infty{\int_0^{\infty}{u^3e^{-u(2k+1)}}du} & & & \color{#D61F06} \text{put }u=\frac{t}{(2k+1)} \end{matrix} \\\Rightarrow I=\displaystyle\sum_{k=0}^\infty{\frac{1}{(2k+1)^4}\int_0^{\infty}{t^3e^{-t}}du}=\displaystyle\sum_{k=0}^\infty{\frac{\Gamma(4)}{(2k+1)^4}}=3!\displaystyle\sum_{k=0}^\infty{\left ( \frac{1}{(2k+1)^4}+\frac{1}{(2k+2)^4}-\frac{1}{(2k+2)^4} \right )} \\\Rightarrow I=6\left ( \displaystyle\sum_{k=1}^\infty{\frac{1}{k^4}} -\displaystyle\sum_{k=1}^\infty{\frac{1}{16k^4}}\right)=6\cdot\zeta(4)\cdot(1-\frac{1}{16})=6\cdot\frac{\pi^4}{90}\cdot\frac{15}{16}=\frac{\pi^4}{16}=\frac{\pi^4}{2^4} \\ \text{note that }\zeta(4)=\frac{\pi^4}{90}$