<s>[INST] A interesting improper integral, $ \int_{0}^1\frac{\ln x}{x^2-x-1}\text{d}x$ $$\displaystyle \int_{0}^1\frac{\ln x}{x^2-x-1}\text{d}x$$
I think there should be a smart way to evaluate this. But I cant see..
 [/INST] Using a partial fraction decomposition, one can write
$$
\frac{1}{x^2-x-1}=\frac{2}{\sqrt{5}}\left[\frac{1}{x-x_+}-\frac{1}{x-x_-}\right],
$$
where $x_{\pm}=\frac{1}{2}\left(1\pm\sqrt{5}\right)$ are the roots of the polynomial $x^2-x-1$.
We now observe that we have the uniformly convergent series expansions
$$
\frac{1}{x-x_+} = -\frac{1}{x_+}\sum_{n=0}^\infty{\left(\frac{x}{x_+}\right)^n},\quad 0<x<1;
$$
$$
\frac{1}{x-x_-} = -\frac{1}{x_-}\sum_{n=0}^\infty{\left(\frac{x}{x_-}\right)^n},\quad 0<x<-x_-;
$$
and
$$
\frac{1}{x-x_-} = \frac{1}{x}\sum_{n=0}^\infty{\left(\frac{x_-}{x}\right)^n},\quad -x_-<x<1.
$$
Putting this together, one obtains
$$
\int_0^1\frac{\log(x)}{x^2-x-1}d\,x = \frac{2}{\sqrt{5}}\times\sum_{n=0}^\infty\left[-\frac{1}{x_+}\int_0^1{\log(x)\left(\frac{x}{x_+}\right)^nd\,x} + \frac{1}{x_-}\int_0^{-x_-}{\log(x)\left(\frac{x}{x_-}\right)^n d\,x} - \int_{-x_-}^1{\frac{1}{x}\log(x)\left(\frac{x_-}{x}\right)^n d\,x}\right].
$$
I'll leave the rest of the computations to the OP, but can expand if necessary.
The final result is $\pi^2/5\sqrt{5}$, as has already been mentioned.
</s>