I know that in Riemann's 1854 paper, he found the following: $$\frac{\ln(\zeta(s))}{s}=\int_1^\infty{J(x)x^{-s-1}dx}\tag{1}$$ Riemann then performed Mellin inversion to get an expression for $J(x)$, and then used Mobius inversion to find an expression for $\pi(x)$
In this video, the following identity is proven: $$\frac{\ln(\zeta(s))}{s}=\int_2^\infty\frac{\pi(x)}{x(x^{s}-1)}dx\tag{2}$$ Is there some type of integral inversion that can be performed on the RHS of $(2)$ to isolate $\pi(x)$? In a similar fashion to how Mellin inversion was done on $(1)$.
They are essentially the same thing:
\begin{aligned} \int_2^\infty{\pi(x)\over x(x^s-1)}\mathrm dx &=\int_0^\infty\pi(x)x^{-1}\cdot{x^{-s}\over1-x^{-s}}\mathrm dx \\ &=\int_0^\infty\pi(x)x^{-1}\sum_{k\ge1}x^{-sk}\mathrm dx \\ &=\sum_{k\ge1}\int_0^\infty\pi(x)x^{-sk-1}\mathrm dx \\ &=\sum_{k\ge1}\int_0^\infty{\pi(x^{1/k})\over k}\cdot x^{-s-1}\mathrm dx \\ &=\int_2^\infty J(x)x^{-s-1}\mathrm dx \end{aligned}