What substitution is ideal here?

$A=\int_0^1\left(\ln 4 +\ln\left(1-\frac{3^x}{4}\right)+2\ln\left(1+3^x\right)\right)dx$ subbing $\frac{3^x}{4}=y\implies dx =\frac{dy}{y \ln3 }$ yields $\begin{aligned} A &=\ln 4+\int_{\frac{1}{4}}^{\frac{3}{4}}\left( \frac{\ln(1-y)}{y}+\frac{2\ln(1+4y)}{y}\right)\frac{dy}{\ln3}\\&=\ln(4)-\frac{1}{\ln3}\left[\operatorname{Li}_{2}(y)+2\operatorname{Li}_{2}(-4y)\right]_{\frac{1}{4}}^{\frac{3}{4}}\\&=\ln4 -\frac{1}{\ln3}\left(\operatorname{Li}_{2}\left(\frac{3}{4}\right)+\operatorname{Li}_{2}\left(\frac{1}{4}\right)-2\operatorname{Li}_{2}(-3)+2\operatorname{Li}_{2}(-1)\right)\\&=\ln 4+ \left(\eta(2)-\ln\left(\frac{1}{4}\right)\ln\left(\frac{3}{4}\right)\right)-2\left(\operatorname {Li}_{2}\left(\frac{3}{4}\right)-\operatorname{Li}_{2}(-3)\right)-\eta(2)\\&= \ln4+\frac{1}{\ln3}\left( \ln(3)\ln(4)-\ln^2(4)+\ln^2(4)\right)\\&=\ln(16)\end{aligned}$ Thus $e^A=16$