Letting $y \mapsto \frac{1}{y}$ we get $\displaystyle I = \int_0^1 \frac{y-1}{(y^3+1)\log{y}}\,\mathrm{d}y$; since $\displaystyle \frac{y-1}{\log{y}}= \int_0^1 y^t \, \mathrm{d}t$, we have
$$\begin{aligned} I & = \int_0^1 \int_0^1 \frac{y^t}{y^3+1}\,\mathrm{d}t\,\mathrm{d}y = \int_0^1 \int_0^1...