Given that $\int_2^x {1\over \log(t)} \, dt = Li(x)$
can be represented as,
$\text{Li(x)}=-\int_{\log(2)}^{\log(x)}\frac{e^{-y}}{y}=\text{Ei}(\log(x))-\text{Ei}(\log(2))$
where $\text{Ei(x)}=\int_{x}^{\infty}\frac{e^{-q}}{q}$ is the Expontential integral,
The simple question is how step by step does ${1\over \log(t)} \,$ become $\frac{e^{-y}}{y}$?
It looks straightforward. $y=-log(t),\ t=e^{-y},\ dt=-e^{-y}dy$. Put it together to get the new expression for Li(x).