Evaluate $ \int_{0}^{1}{\frac{\ln{x}}{x^2-x-1}dx}$

For what it's worth, here are some computations to help you get further from where you stopped.

For $|x_0|>1$, which is the case of your $x_0=\frac{1+\sqrt{5}}{2}$, we have: $$ \int_0^1\frac{\ln x}{x-x_0}dx=-\frac{1}{x_0}\int_0^1\frac{\ln x}{1-\frac{x}{x_0}}dx=-\frac{1}{x_0}\int_0^1\ln x\sum_{k\geq 0}\left(\frac{x}{x_0}\right)^kdx $$ $$ =-\frac{1}{x_0}\sum_{k\geq 0}\frac{1}{x_0^k}\int_0^1x^k\ln x dx=\sum_{k\geq 0}\frac{x_0^{-(k+1)}}{(k+1)^2}. $$ Now a power series computation starting from $(1-y)^{-1}=\sum_{k\geq 0}y^k$ yields $$ \sum_{k\geq 0}\frac{y^{k+1}}{(k+1)^2}=-\int_0^y\frac{\ln(1-t)}{t}dt=\mathrm{Li}_2(y) $$ the dilogarithm. So $$ \int_0^1\frac{\ln x}{x-x_0}dx=\mathrm{Li}_2\left(\frac{1}{x_0}\right)\quad\forall\;|x_0|>1. $$

I think I'll let you handle the case $-1\leq x_1\leq 1$.