$\displaystyle \int_{0}^1\frac{\ln x}{x^2-x-1}\text{d}x$
I think there should be a smart way to evaluate this. But I cant see..
$\displaystyle \int_{0}^1\frac{\ln x}{x^2-x-1}\text{d}x$
I think there should be a smart way to evaluate this. But I cant see..
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
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.