\begin{align} &\quad\int_0^1\frac{x-1}{(1+x)\ln x}\mathrm{~d}x\\ &\xlongequal{\ln x\to-x}\int_0^\infty\frac{\mathrm{e}^{-x}-\mathrm{e}^{-2x}}{\left(1+\mathrm{e}^{-x}\right)x}\mathrm{~d}x\\ &=\int_0^\infty\frac{\mathrm{e}^{-x}-\mathrm{e}^{-2x}}{x}\sum_{n=0}^\infty(-)^n\mathrm{e}^{-nx}\mathrm{~d}x\\ &=\sum_{n=0}^\infty(-)^n\underbrace{\int_0^\infty\frac{\mathrm{e}^{-(n+1)x}-\mathrm{e}^{-(n+2)x}}{x}\mathrm{~d}x}_{\text{Frullani}}\\ &=\sum_{n=0}^\infty(-)^n\ln\left(\frac{n+2}{n+1}\right)\\ &=\sum_{n=0}^\infty\ln\left(\frac{2n+2}{2n+1}\cdot\frac{2n+2}{2n+3}\right)\\ &=\ln\left(\prod_{n=0}^\infty\frac{(2n+2)^2}{(2n+1)(2n+3)}\right)\\ &=\underbrace{\lim_{n\to\infty}\ln\left\{\frac{[(2n)!!]^2(2n+1)}{[(2n+1)!!]^2}\right\}}_{\text{Wallis}}\\ &=\ln\left(\frac{\pi}{2}\right). \end{align} 下面是三个相关的结论(证明有时间再写)：1.\int_{0}^{1}\frac{x^\alpha-1}{\mathrm {ln}\,x}\mathrm{d}x=\mathrm{ln}\,(\alpha+1) 2.\int_{0}^{1}\frac{x^\alpha-1}{(x+1)\mathrm {ln}\,x}\mathrm{d}x=\mathrm{ln}\,(\frac{\Gamma(\frac{\alpha}{2}+1)}{\Gamma(\frac{\alpha+1}{2})}\sqrt{\pi}) 3.\int_{0}^{1}\frac{(x-1)^2}{(x^n-1)\,\mathrm{ln}\,x}\mathrm{d}x=\mathrm{ln}\,(\frac{\Gamma(\frac{3}{n})\Gamma(\frac{1}{n})}{\Gamma^2(\frac{2}{n})}) 注1：对于1，利用含参变量积分来证明。注2：对于2和3，都可以用 \psi(x) （Digamma函数）的积分形式来证明注3：对于3，特别地，当 n=2 时积分值为 \mathrm{ln}\,(\frac{\pi}{2}) ,当 n=4 时，积分值为 \mathrm{ln}\,(\sqrt{2}) .}