Consider a function $$J(x)=\sum_{n}\frac{\pi(\sqrt[n]{x})}{n}$$ with the prime counting function $\pi(x)$. This sum is finite, because for $n$ big enough $\sqrt[n]{x}<2$ and therefore $\pi(\sqrt[n]{x})=0$. The author of a book then applies the möbius inversion formula to get $$\pi(x)=\sum_n\mu(n)\frac{J(\sqrt[n]{x})}{n}$$ However, I am only aware of the inversion formula in for cases like $$g(x)=\sum_{d\mid x}f(d)$$ or $$g(x)=\sum_{n=1}^x f\Big(\frac{x}{n}\Big)$$ Also I was unable to adapt the known proofs for the other results due to technical reasons. Especially because of the upper bound of summation (the first $n$ with $\sqrt[n]{x}<2$) makes it hard to juggle with sums embedded into each other and makes it impossible (for me) to copy the quite easy proofs of the other cases.
I would appreciate any help on this problem how to prove the wanted version of the inversion formula
Notice that
\begin{aligned} \sum_{n\ge1}\mu(n){J(\sqrt[n]x)\over n} &=\sum_{n\ge1}{\mu(n)\over n}\sum_{m\ge1}{\pi(\sqrt[mn]x)\over m} =\sum_{n\ge1}\sum_{m\ge1}\mu(n){\pi(\sqrt[mn]x)\over mn} \\ &=\sum_{k\ge1}{\pi(\sqrt[k]x)\over k}\sum_{mn=k}\mu(n)=\pi(x), \end{aligned}
where the last equality follows from the fact that
$$ \sum_{mn=k}\mu(n)=\sum_{d|k}\mu(d)=\begin{cases} 1 & k=1, \\ 0 & k>1. \end{cases} $$