How to solve this definite integral with involving logarithm: $\int_0^1\left( \ln (4-3^x)+2 \ln(1+3^x)\right)dx$ $$\int_0^1\left( \ln (4-3^x)+2 \ln(1+3^x)\right)dx.$$
The answer given $\log 16$. Can anyone explain this question for me ? Thanks a lot. 
 A: This integral involves the use of Polylogarithms. Recall that:$$\int-\frac{\ln(1-x)}{x}dx=\mathrm{Li}_2(x)+C$$
For your case we will need to apply the definition given above. We have:$$\int_{0}^{1} \ln (4-3^x)+2 \ln(1+3^x)dx=\\\int_{0}^{1} \ln (4-3^x)+ \int_0^12 \ln(1+3^x)dx$$Now let's take a look at the first one:$$\\ \int_{0}^{1} \ln \left(4\left(1-\frac{3^x}{4}\right)\right)dx=\\ \int_{0}^{1} \ln (4)+\ln \left(1-\frac{3^x}{4}\right)dx=\\$$$$\bbox[5px,border:2px solid red]
{u=\frac{3^x}{4}\quad \quad du=\ln(3)u\ dx}$$$$\\ \int_{\frac14}^{\frac{3}{4}} \ln (4)+\frac{\ln(1-u)}{\ln(3)u}du=\\ \left[x\ln(4)-\frac{\mathrm{Li}_2(\frac{3^x}{4})}{\ln(3)}\right]_0^1$$
Now the second one: $$\int_0^12 \ln(1+3^x)dx=\\$$$$\bbox[5px,border:2px solid red]
{z=-3^x\quad \quad dz=\ln(3)z}$$$$\\ 2\int_1^{-3}\frac{\ln(1-z)}{\ln(3)z}dz=\\\left[\frac{-2\mathrm{Li}_2(-3^x)}{\ln(3)}\right]_0^1$$
Therefore, we finally have: $$\left[\frac{x\ln(4)\ln(3)-\mathrm{Li}_2(\frac{3^x}{4})-\mathrm{Li}_2(-3^x)}{\ln(3)}\right]_0^1=\frac{\ln(3)\ln(4)-\mathrm{Li}_2(-3)-\mathrm{Li}_2(\frac{3}{4})-\frac{\pi^2}{6}+\mathrm{Li}_2(\frac{1}{4})}{\ln(3)} \approx$$$$\bbox[5px,border:2px solid black]
{2.773}$$
