Can anyone please determine integral below?

Question

I was creating a paper on P.N.T but I stucked here so,

Answer 1

From the point of view of distributions we may add the jumps at all the prime numbers $p$ to get : $$R'(t)=\sum_{p\;\text{prime}}\left[\frac{\ln(p)}{p}\delta(t-p)\right]-\frac 1t$$ (since a term $\dfrac{\ln(p)}{p}$ is added to $R(t)$ each time $t$ 'crosses a prime $p$')

We may then conclude that (restraining the integral to the primes $\,p\le x$) :

\begin{align} \int_0^x \frac {t\,R'(t)}{\ln(t)}dt&=\sum_{p\le x}\left[\int_0^x\frac{\ln(p)}{p}\frac t{\ln(t)}\delta(t-p)\,dt\right]-\int_0^x\frac t{t\,\ln(t)} dt\\ &=\sum_{p\le x}\frac{\ln(p)}{p}\frac {p}{\ln(p)}-\operatorname{li}(x)\\ \\ \int_0^x \frac {t\,R'(t)}{\ln(t)}dt&=\pi(x)-\operatorname{li}(x)\\ \end{align} from the definition of the prime-counting function $\pi(x)$ and the logarithmic integral $\;\operatorname{li}(x)$.