Evaluation of $\int_{0}^{1}\frac{\ln x}{x^2-x-1}dx$ 
Evaluation of $\displaystyle \int_{0}^{1}\frac{\ln x}{x^2-x-1}dx$

$\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{\infty}\frac{\ln x}{x^2-x-1}dx=\int_{0}^{\infty}\frac{\ln(x)}{\left(x-\frac{1}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2}dx$
Now Put $\displaystyle \left(x-\frac{1}{2}\right)=\frac{\sqrt{5}}{2}\sec \theta\;,$ Then $\displaystyle dx = \frac{\sqrt{5}}{2}\sec \theta \tan \theta$
So $$ I= -\frac{2}{\sqrt{5}}\int_{-\frac{1}{2}}^{\infty}\frac{\ln(\sqrt{5}\sec \theta+1)-\ln(2)}{\tan \theta}\cdot \sec \theta d\theta$$
So $$I = \frac{2}{\sqrt{5}}\int_{-\frac{1}{2}}^{\infty}\frac{\ln(2)+\ln(\cos \theta)-\ln(\sqrt{5}+\cos \theta)}{\sin \theta} d\theta$$
Now How can I solve after that, Help Required, Thanks
 A: So, by the comment of the OP, the integral we have to study is $$\int_{0}^{1}\frac{\log\left(x\right)}{x^{2}-x-1}dx.$$ We note that $$\begin{align} I= & \int_{0}^{1}\frac{\log\left(x\right)}{x^{2}-x-1}dx \\ = & \int_{0}^{1}\frac{\log\left(x\right)}{\left(x-\frac{1-\sqrt{5}}{2}\right)\left(x-\frac{1+\sqrt{5}}{2}\right)}dx \\ = & -\frac{2}{\sqrt{5}}\left(\int_{0}^{1}\frac{\log\left(x\right)}{2x+\sqrt{5}-1}dx-\int_{0}^{1}\frac{\log\left(x\right)}{2x-\sqrt{5}-1}dx\right) \\ = & -\frac{2}{\sqrt{5}}\left(I_{1}-I_{2}\right), \end{align}$$ say. Let us analyze $I_{1}$. Integrating by part we get $$I_{1}=\int_{0}^{1}\frac{\log\left(x\right)}{2x+\sqrt{5}-1}dx=\frac{1}{\sqrt{5}-1}\int_{0}^{1}\frac{\log\left(x\right)}{\frac{2x}{\sqrt{5}-1}+1}dx=-\frac{1}{2}\int_{0}^{1}\frac{\log\left(\frac{2x}{\sqrt{5}-1}+1\right)}{x}dx
 $$ $$=\frac{1}{2}\textrm{Li}_{2}\left(-\frac{2}{\sqrt{5}-1}\right)
 $$ where $\textrm{Li}_{2}(x)$ is the Dilogarithm function. In a similar way we can find that $$I_{2}=\int_{0}^{1}\frac{\log\left(x\right)}{2x-\sqrt{5}-1}dx=\frac{1}{2}\textrm{Li}_{2}\left(\frac{2}{\sqrt{5}+1}\right)
 $$ so $$I=\frac{\textrm{Li}_{2}\left(\frac{2}{\sqrt{5}+1}\right)-\textrm{Li}_{2}\left(-\frac{2}{\sqrt{5}-1}\right)}{\sqrt{5}}
 $$ and since holds $$\textrm{Li}_{2}\left(\phi^{-1}\right)=\frac{1}{10}\pi^{2}-\log^{2}\left(\phi\right),\,\textrm{Li}_{2}\left(-\left(\phi-1\right)^{-1}\right)=-\frac{1}{10}\pi^{2}-\log^{2}\left(\phi\right)
 $$ where $\phi=\frac{\sqrt{5}+1}{2}
 $ is the golden ratio we have $$I=\frac{\pi^{2}}{5\sqrt{5}}.$$ 
