一个看似简单的积分
\[\Large\displaystyle \int_{0}^{1}\left [ \frac{1+\sqrt{1-x}}{x} +\frac{2}{\ln\left ( 1-x \right )}\right ]\mathrm{d}x
\]
\(\Large\mathbf{Solution:}\)
方法一：令\(\ln(1-x)=-t\)，有
\[\begin{align*}
\int_{0}^{1}\left [ \frac{1+\sqrt{1-x}}{x} +\frac{2}{\ln\left ( 1-x \right )}\right ]\mathrm{d}x =& \int_{0}^{\infty }e^{-t}\left ( \frac{1}{1-e^{-\frac{t}{2}}}-\frac{2}{t} \right )\mathrm{d}t\\
=&\int_{0}^{\infty }\left [ -2\left ( \frac{e^{-t}}{t}-\frac{1}{e^{t}-1} \right )-\frac{e^{-\frac{t}{2}}}{1+e^{\frac{t}{2}}} \right ]\mathrm{d}t\\=&-2\int_{0}^{\infty }\left ( \frac{e^{-t}}{t} -\frac{1}{e^{t}-1}\right )\mathrm{d}t-\int_{0}^{\infty }\frac{e^{-\frac{t}{2}}}{1+e^{\frac{t}{2}}}\mathrm{d}t\\
=&2\gamma -\int_{0}^{\infty }\left ( e^{-\frac{t}{2}}-\frac{e^{-\frac{t}{2}}}{1+e^{-\frac{t}{2}}} \right )\mathrm{d}t\\=&\Large\boxed{\displaystyle \color{blue}{2\gamma -2+2\ln 2}}
\end{align*}\]
方法二：令\(1-x\rightarrow x\)，有
\[\begin{align*}
\int_{0}^{1}\left [ \frac{1+\sqrt{1-x}}{x} +\frac{2}{\ln\left ( 1-x \right )}\right ]\mathrm{d}x =& \int_{0}^{1}\left ( \frac{1+\sqrt{x}}{1-x}-\frac{2}{\ln x} \right )\mathrm{d}x\\
=&\int_{\infty }^{0}\int_{0}^{1}\frac{x^{t}\ln x}{1-x}+\frac{x^{t+\frac{1}{2}}\ln x}{1-x}+2x^{t}\mathrm{d}x\mathrm{d}t\\=&-\int_{0}^{\infty }\left [\psi _{1}\left ( t+1 \right )+\psi _{1}\left ( t+\frac{2}{3} \right )-\frac{2}{t+1}\right ]\mathrm{d}t\\
=&-\psi _{0}\left ( 1 \right )-\psi _{0}\left ( \frac{3}{2} \right )\\=&\Large\boxed{\displaystyle \color{blue}{2\gamma -2+2\ln 2}}
\end{align*}\]
