The original problem is
$$\int_{0}^{1} \frac{1-x}{\ln x}(x + x^2 + x^{2^2}+ x^{2^3}+\cdots) dx$$
Now the problem has been solved through the tips from Mr. Nosrati. The solution is as follows:
\begin{align} \int_{0}^{1}\frac{1-x}{\ln x}(x+x^2+x^{2^2}+\cdots) \, dx &= \int_{0}^{1}\frac{1-x}{\ln x} \, \sum_{n=0}^{+\infty}x^{2^n} \, dx \\ &= -\sum_{n=0}^{+\infty}\int_{0}^{+\infty}\frac{1-e^{-u}}{u}e^{-(2^n+1)u} \, du \\ &= -\sum_{n=0}^{+\infty}\int_{0}^{+\infty}\frac{e^{-(2^n+1)u}-e^{-(2^n+2)u}}{u} \, du \\ &= \sum_{n=0}^{\infty} \ln\frac{2^n+1}{2^n+2}(e^0-\lim_{x\to+\infty}e^{-x}) \\ &= \sum_{n=0}^{\infty} \ln\frac{1+2^{-n}}{1+2^{-(n+1)}} \\ &= -\ln 3. \end{align}
And he put forward a new challenging question,that is to calculate
$$\prod_{k=0}^{\infty}\frac{2^k+1}{2^k+2}$$
I have made a rough estimate of it as follows:
$$\frac{1}{2e^{\frac{\pi^2}{6}}}<\prod_{k=0}^{n}\frac{k^2+1}{k^2+2}<\frac{1}{2}$$
which is given by
$$\prod_{i=1}^{n}\frac{k^2+2}{k^2+1}=\prod_{i=1}^{n}(1+\frac{1}{k^2+1})< \prod_{i=1}^{n}e^{\frac{1}{k^2+1}}<\prod_{i=1}^{n}e^{\frac{1}{k^2}}=e^{\sum_{i=1}^{n}\frac{1}{k^2}}<e^{\frac{\pi^2}{6}}$$