此题由tian275461提供。题目非常困难，仅供观赏(水平一般的同学请勿模仿 )。

求积分
\[ I=\int_{0}^{1}{\frac{\ln{x}}{x^2-x-1}dx} \]
解（tian275461）

由方程$x^2-x-1=0$ 的两个根，为了简单起见，我们记$ r_{1}=\varphi=\dfrac{1+\sqrt{5}}{2},r_{2}=\dfrac{1-\sqrt{5}}{2}=1-\varphi$,
且$r_{1}-r_{2}=\sqrt{5},\varphi^{2}=\varphi+1 , \dfrac{\varphi-1}{\varphi}=\dfrac{1}{\varphi^{2}} $
则有

\[ \begin{align*}
I&=\int_{0}^{1}{\frac{\ln{x}}{x^2-x-1}dx}\\
&=\frac{1}{r_{1}-r_{2}}\int_{0}^{1}{\ln{x}\left(\frac{1}{x-\varphi}-\frac{1}{x-(1-\varphi)} \right)dx}\\
&=\dfrac{1}{\sqrt{5}}\int_{0}^{1}{\ln{x}\left(\frac{1}{x-\varphi}-\frac{\varphi}{\varphi x+1}\right)dx}\\
&=\dfrac{1}{\sqrt{5}}\int_{0}^{1}{\frac{\ln{x}}{x-\varphi}dx}-\frac{\varphi}{\sqrt{5}}\int_{0}^{1}{\frac{\ln{y}}{\varphi y+1}dy}\\
&=\frac{1}{\sqrt{5}}\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{\varphi u}}{u-1}du}-\frac{1}{\sqrt{5}}\int_{0}^{\varphi}{\frac{\ln{\frac{u}{\varphi}}}{u+1}du}\\
&=\frac{\ln{\varphi}}{\sqrt{5}}\left(\int_{0}^{\frac{1}{\varphi}}{\frac{1}{u-1}du}+\int_{0}^{\varphi}{\frac{1}{u+1}du} \right)-\frac{1}{\sqrt{5}}\left(\int_{0}^{\varphi}{\frac{\ln{u}}{1+u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du} \right)\\
&=\frac{\ln{\varphi}}{\sqrt{5}}\ln{\frac{\varphi^{2}-1}{\varphi}}-\frac{1}{\sqrt{5}}\left(\int_{0}^{\varphi}{\frac{\ln{u}}{1+u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du} \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{0}^{\varphi}{\frac{\ln{u}}{1+u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du} \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{1}^{1+\varphi}{\frac{\ln{(u-1)}}{u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}  \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{1}^{\varphi^{2}}{\frac{\ln{(u-1)}}{u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}  \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{\frac{1}{\varphi^{2}}}^{1}{\frac{\ln{(1-u)}-\ln{u}}{u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}  \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}-\ln{(1-u)}}{u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}  \right)\\
&=-\frac{2}{\sqrt{5}}\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}+\frac{1}{\sqrt{5}}\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{(1-u)}}{1-u}du}\\
&=-\frac{2}{\sqrt{5}}\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=-\frac{2}{\sqrt{5}}\int_{\frac{1}{\varphi^{2}}}^{1}{\frac{\ln{(1-u)}}{u}du}-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=-\frac{2}{\sqrt{5}}\left(\int_{0}^{1}{\frac{\ln{(1-u)}}{u}du}-\int_{0}^{\frac{1}{\varphi^{2}}}{\frac{\ln{(1-u)}}{u}du} \right)-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=\frac{\pi^{2}}{3\sqrt{5}}-\frac{2}{\sqrt{5}}\Li_{2}{\left(\frac{1}{\varphi^{2}}\right)}-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=\frac{\pi^{2}}{3\sqrt{5}}-\frac{2}{\sqrt{5}}\left(\frac{\pi^{2}}{15}-\ln^{2}{\varphi}
\right)-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=\frac{\pi^{2}}{5\sqrt{5}}
\end{align*} \]

$\square$

最后一步用了dilogarithm函数的性质，详细可以参见http://mathworld.wolfram.com/Dilogarithm.html

______________________________________________________________________________

简单一点的做法

我们有
\[ \Li_{2}(x)=-\int_{0}^{x}\frac{\ln(1-t)}{t}dt \]
则
\begin{align*}
\int_{0}^{1}\frac{\ln{x}}{x-a}&=-\frac{1}{a}\cdot\int_{0}^{1}\frac{\ln{x}}{1-\frac{x}{a}}dx\\
&= -\frac{1}{a}\cdot\left(\int_{0}^{1}\frac{\ln\left(\frac{x}{a}\right)}{1-\frac{x}{a}}dx+\ln{a}\int_{0}^{1}\frac{1}{1-\frac{x}{a}}dx\right)\\
&=-\int_{0}^{\frac{1}{a}}\frac{\ln{t}}{1-t}dt-\frac{\ln{a}}{a}\int_{0}^{1}\frac{1}{1-\frac{x}{a}}dx\\
&=-\int_{0}^{\frac{1}{a}}\frac{\ln(1-t)}{t}dt\\
&=\Li_{2}\left(\frac{1}{a}\right)
\end{align*}
注意到
\[ x^2-x-1=\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right) \]
\[ I=\int_{0}^{1}\frac{\ln{x}}{x^2-x-1}dx=\frac{1}{\sqrt{5}}\left(\int_{0}^{1}\frac{\ln{x}}{x-\frac{\sqrt{5}+1}{2}}dx-\int_{0}^{1}\frac{\ln{x}}{x-\frac{1-\sqrt{5}}{2}}dx \right)=\frac{1}{\sqrt{5}}\left(\Li_{2}\left(\frac{\sqrt{5}-1}{2}\right)-\Li_{2}\left(-\frac{\sqrt{5}+1}{2}\right)\right)\]
记$\phi=\frac{\sqrt{5}+1}{2}$,则
\[ I=\frac{1}{\sqrt{5}}\left(\Li_{2}\left(\frac{1}{\phi}\right)-\Li_{2}(-\phi)\right) \]
我们还知道公式
\[ \Li_{2}\left(\frac{1}{\phi}\right)=\frac{1}{10}\pi^2-\ln^{2}\phi \]
\[ \Li_{2}(-\phi)=-\frac{1}{10}\pi^2-\ln^2\phi \]
则
\[ I=\frac{1}{5\sqrt{5}}\pi^2 \]