Given that $$\int_{1/\phi}^{1/\phi^2}{ \dfrac{\ln(1-x)}{x}}dx=\dfrac{\pi^2}{30}$$ Find the value of $$\int_{1/\phi}^{1/\phi^2} \left(\dfrac{\ln(1-x)}{x}\right)^2 dx$$ in terms of $\phi$ and $\pi$. Where $\phi=\frac{1+\sqrt 5}{2}$ is the golden ratio.
I have tried to do it by taylor series, also tried integration by parts but it is getting ugly and too many terms are coming.
Source: Made by Prof. Raghava.
We just need to integrate by parts, using $\left(1-\frac{1}{x}\right)'=\frac{1}{x^2}$ we get: $$\require{cancel} \int_\frac{1}{\phi}^\frac{1}{\phi^2}\frac{\ln^2(1-x)}{x^2}dx= \left(1-\frac{1}{x}\right)\ln^2 (1-x)\bigg|_\frac{1}{\phi}^\frac{1}{\phi^2}-2\int_\frac{1}{\phi}^\frac{1}{\phi^2}\frac{\cancel{1-x}}{x}\frac{\ln (1-x)}{\cancel{1-x}}dx$$ $$=\left(\frac{4}{\phi}-\phi\right)\ln^2 \phi-2\underbrace{\int_\frac{1}{\phi}^\frac{1}{\phi^2}\frac{\ln (1-x)}{x}dx}_{=\frac{\pi^2}{30}}=\boxed{\left(\frac{4}{\phi}-\phi\right)\ln^2 \phi-\frac{\pi^2}{15}}$$ The reason why the more expected $\left(-\frac{1}{x}\right)'=\frac{1}{x^2}$ wasn't used in order to integrate by parts is that indeed too many terms are coming out (as OP noticed) and we would need some extra steps to do, involving partial fraction, but this way we can cancel the additional $1-x$ term that comes out after differentiating $\ln^2(1-x)$.