Let: $$T_n=\int_0^1\int_0^t \frac{1}{x(1-x)}\frac{x^{2^n}}{x^{2^n}+1}\,\mathrm{d}x\,\mathrm{d}t$$
Through experimentation it looks like $T_n=\log 2/2^n$, so we should have the relation $2T_n=T_{n+1}$ and $T_0=\log 2$. However I having trouble actually proving this, the substitution $u=2x$ doesn't seem to work. I'm not hugely familiar with double-integration so perhaps that is where I am going wrong. I would love some help!
$\textbf{Hint}$: Since the integrand has no $t$ dependence, we can change the order of integration to simplify the problem like so
$$= \int_0^1\int_x^1 \frac{1}{x(1-x)}\frac{x^{2^n}}{x^{2^n}+1}\,\mathrm{d}t\,\mathrm{d}x = \int_0^1 \frac{x^{2^n-1}}{x^{2^n}+1}\mathrm{d}x$$