Generalization Of A Problem = Unique Symmetry

$\begin{aligned} I(n) & = \int_0^1 \frac {\color{#3D99F6}{\ln(1-x^n)}}x dx & \small \color{#3D99F6}{\text{By Maclaurin's series,}} \\ & = \int_0^1 \frac 1x \left( \color{#3D99F6}{-\sum_{k=1}^\infty \frac {x^{nk}}k} \right) dx \\ & = - \int_0^1 \sum_{k=1}^\infty \frac {x^{nk-1}}k dx \\ & = - \sum_{k=1}^\infty \frac {x^{nk}}{nk^2} \bigg|_0^1 \\ & = - \sum_{k=1}^\infty \frac 1{nk^2} \\ & = - \frac {\color{#3D99F6}{\zeta(2)}}n & \small \color{#3D99F6}{\zeta(\cdot) \text{ is the Riemann zeta function}} \\ & = - \frac {\color{#3D99F6}{\pi^2}}{\color{#3D99F6}{6}n} \end{aligned}$

Therefore, $\displaystyle \left \lfloor \left| \sum_{n=1}^\infty \frac {I(n)}n \right| \right \rfloor = \left \lfloor \left| \sum_{n=1}^\infty - \frac {\pi^2}{6n^2} \right| \right \rfloor = \left \lfloor \left|- \left(\frac {\pi^2}6\right)^2 \right| \right \rfloor = \boxed{2}$