How to calculate the following integral?
$$\int_0^1\frac{\ln x}{x^2-x-1}\mathrm{d}x=\frac{\pi^2}{5\sqrt{5}}$$
How to calculate the following integral?
$$\int_0^1\frac{\ln x}{x^2-x-1}\mathrm{d}x=\frac{\pi^2}{5\sqrt{5}}$$
You can start writing $$x^2-x-1=(x-r_1)(x-r_2)$$ where $r_{1,2}=\frac{1}{2} \left(1\pm\sqrt{5}\right)$ and use partiel fraction decomposition. So,$$\frac 1{x^2-x-1}=\frac{1}{{r_2}-{r_1}} \Big(\frac{1}{x-r_2}-\frac{1}{x-r_1}\Big)$$ and use $$\int \frac{\log(x)}{x+a}=\text{Li}_2\left(-\frac{x}{a}\right)+\log (x) \log \left(1+\frac{x}{a}\right)$$