On geometric series and the logarithmic integral

With this I'm recycling a previous deleted post that contained a serious mistake (in my calculation a geometric series was not convergent). Also was interested in a diffrent question I am asking it now.

Seems interesting do specizalitions, and after combine with asymptotics, Chirita, PROBLEMA 242 from La Gaceta de la RSME, Vol. 18.1 (2015), pages 124-125, (in spanish). One can write $$\sum_{p\leq x } \operatorname{Li} \left( x^p \right)=\operatorname{Li} \left( x\right)\pi(x) +\vartheta(x)+\sum_{p\leq x} \int_1^x\frac{t^{p-1}}{\log t}dt ,$$ where $p$ denote prime number, $\pi(x)$ the prime-counting function and $\vartheta(x)$ is the first Chebyshev function.

Prime Number Theorem implies that the following asymptotic equivalence holds $$\frac{1}{x}\sum_{p\leq x } \operatorname{Li} \left( x^p \right)\sim 1+\frac{x}{\log^2 x}+\frac{1}{x}\sum_{p\leq x} \int_1^x\frac{t^{p-1}}{\log t}dt,$$ as $x\to\infty$. (I don't know if it's possible do better calculations that mine in this step to get interesting asymptotic inequalitites, if you want you can add a comment doing feedback about it, ut my main question is the following

Question. Can you provide me hints to study the asymptotic behaviour of $$\sum_{\substack{p\leq x\\p\text{ prime}}} \int_1^x\frac{t^{p-1}}{\log t}dt$$ as $x\to\infty$? Many thanks.

We know that $$\int \frac{y^{m-1}}{\log y}dy=\operatorname{Ei}(m\log y)+\text{ constant},$$ where with $\operatorname{Ei}(x)$ we are denoting the exponential interal $\operatorname{Ei}$. And I know summation methods.