I'm reading the Riemann's article on the number of prime numbers with a detailed explanation by W. Dittrich, and stucked to prove the Riemann's final formula
$$ J(x) := \sum_{p^n \leq x} \dfrac{1}{n} = \operatorname{li}(x) - \sum_{\rho} \operatorname{li}(x^\rho) + \int_{x}^{\infty} \dfrac{1}{t^2-1} \dfrac{dt}{t\log t} -\log 2, \quad \forall x>1 $$
using $$ \log\zeta(s) = \dfrac{s}{2}\log\pi - \log 2 - \log(s-1) - \log\Gamma\left(1+\dfrac{s}{2}\right) + \sum_{\rho}\log\left( 1-\dfrac{s}{\rho} \right) $$ and $$ -\log\Gamma \left( 1+\dfrac{s}{2} \right) = \log\dfrac{2}{s}\dfrac{1}{\Gamma(s/2)}= \lim_{m\to\infty} \left[ \sum_{n=1}^{m} \log\left( 1+\dfrac{s}{2n} \right) - \dfrac{s}{2}\log m \right], $$ where $J(x)$, which is the weighted prime counting function, and $\log\zeta(s)$ are connected with the inverse Mellin transform $$ J(x) = \dfrac{1}{2\pi i} \int_{a-i\infty}^{a+i\infty} \dfrac{\log \zeta(s)}{s} x^s ds, \quad a>1. $$
Riemann, in his paper, considered the integration by parts below $$ \int_{a-i\infty}^{a+i\infty} \dfrac{\log \zeta(s)}{s} x^s ds = \dfrac{-1}{\log x} \int_{a-i\infty}^{a+i\infty} \dfrac{ d\dfrac{\log \zeta(s)}{s}} {ds} x^s ds $$ and concluded that it suffices to evaluate the integral with the form $$ \int_{a-i\infty}^{a+i\infty} \dfrac{1}{\log x} \dfrac{ d \left[ \dfrac{1}{s} \log \left( 1-\dfrac{s}{\beta} \right) \right] } {ds} x^s ds. $$
I stucked here to evaluate this integral. In Riemann's paper, he used
$$ \dfrac{ d \left[ \dfrac{1}{s} \log \left( 1-\dfrac{s}{\beta} \right) \right] } {d\beta} = \dfrac{1}{(\beta-s)\beta} $$ and established the result $$ \int_{a-i\infty}^{a+i\infty} \dfrac{1}{\log x} \dfrac{ d \left[ \dfrac{1}{s} \log \left( 1-\dfrac{s}{\beta} \right) \right] } {ds} x^s ds = \int_{\infty}^{x} \dfrac{x^{\beta-1}}{\log x} dx \quad \text{or} \quad \int_{0}^{x} \dfrac{x^{\beta-1}}{\log x} dx $$ depending on the sign of $\operatorname{Re}(\beta)$, and he constructed the first formula using $\operatorname{li}(x)$.
From here, I have no idea about his derivation and I also wonder that where from $dx$ comes instead of $ds$. I am aware of the simpler derivation by von Mangoldt using the Chevichev function $\psi(x)$, but I want to go through this direct calculation. Can this integration be explicitly converted into the final formula using $\operatorname{li}(x)$?
Any hints or suggestions are appreciated. Or other links with more detailed explanation in this integration will be helpful very much. Thanks for all comments and answers!