I would like to show that
$ I_{n}=\int_0^1 \frac{x^n \ln x}{x-1}\mathrm dx \sim_{n\rightarrow \infty} \frac{1}{n}$
Using the change of variable $u=x^n$:
$ I_{n}=\frac{1}{n^2} \int_0^1 \frac{u^{1/n} \ln u}{u^{1/n}-1} \mathrm du=\frac{1}{n^2}\left(\int_0^1 \ln x \mathrm dx+\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx \right)=\frac{-1}{n^2}+\frac{1}{n^2}\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx=o(1/n)+\frac{1}{n^2}\int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx$
So I have to show that
$ \int_0^1 \frac{\ln x}{x^{1/n}-1} \mathrm dx \sim_{n \rightarrow \infty} n$
Could you help me?