We want to obtain $\displaystyle\int_0^1\frac{\ln(x)}{x-1}dx=\int_0^\infty\frac{v}{e^v-1}dv$. The data of our substitution is
$$u=\ln x\iff x=e^u \\ du=\frac{dx}{x}\iff dx=e^udu \\ x=0\iff u=-\infty \\ x=1 \iff u=0$$
So we have
$$\int_0^1\frac{\ln(x)}{x-1}dx=\int_{-\infty}^0\frac{u}{e^u-1}(e^udu).$$
Using the substitution $v=-u$, we have
$$\int_{-\infty}^0\frac{u}{e^u-1}(e^udu)=\int_{\infty}^0\frac{-v}{e^{-v}-1}e^{-v}(-dv)=\int_0^\infty\frac{v}{e^v-1}dv.$$
