TITLE: Evaluate $ \int_{0}^{1}{\frac{\ln{x}}{x^2-x-1}dx}$
QUESTION [11 upvotes]: How can we evaluate the definite integral
$$\displaystyle \int_{0}^{1}{\frac{\ln{x}}{x^2-x-1}dx}$$
I tried many times but still have no idea.

REPLY [5 votes]: 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 $-1\leq x_1\leq 1$.