Question about Selberg's formula Let p,q be prime. 
Is it true that 
$$\sum_{p\leq x}(\log p)^2 \sim x \log x~~~?\hspace{5mm}(1)$$
I haven't been able to support modest numerical evidence with the little I know of relations involving sums of logs.   
The reason I wonder is that if this is true then 
$$\sum_{pq \leq x}\log p\log q \sim x \log x\hspace{5mm}(2) $$ by Selberg's relation.*    
I guess it would also be true that $$\sum_{p\leq n}(\log p)^2 \sim n\log n\sim p_n.$$
This has possibly been asked before but a quick search here didn't turn up anything. It may not be true!
Thanks for any insights, hints, references.
*Selberg: $\sum_{p \leq x}(\log p)^2+ \sum_{pq \leq x}\log p\log q = 2x\log x + O(x)$
 A: For some $\eta\in(n,n+1)$,
$$
\begin{align}
&\sum_{n\le x}\pi(n)(\log(n+1)^2-\log(n)^2)\\
&=\sum_{n\le x}\pi(n)\frac{2\log(\eta)}{\eta}\\
&=\sum_{n\le x}\left(\frac{n}{\log(n)}+O\left(\frac{n}{\log(n)^2}\right)\right)\left(\frac{2\log(n)}{n}+O\left(\frac{\log(n)}{n^2}\right)\right)\\
&=2x+O\left(\frac{x}{\log(x)}\right)
\end{align}
$$
Therefore, summing by parts,
$$
\begin{align}
\sum_{p\le x}\log(p)^2
&=\sum_{n\le x}\log(n)^2(\pi(n)-\pi(n-1))\\[6pt]
&=\log(x)^2\pi(x)-\sum_{n\le x}\pi(n)(\log(n+1)^2-\log(n)^2)\\
&=\log(x)^2\left(\frac{x}{\log(x)}+\frac{x}{\log(x)^2}+O\left(\frac{x}{\log(x)^3}\right)\right)-2x+O\left(\frac{x}{\log(x)}\right)\\
&=x\log(x)-x+O\left(\frac{x}{\log(x)}\right)
\end{align}
$$

Experimental Results
$$
\begin{array}{r|c|c|c}
x&\sum_{p\le x}\log(p)^2&x\log(x)-x&\frac{x}{\log(x)}\\
\hline\\
100 & 309.0926254 & 360.5170186 & 21.71472410\\
1000 & 5686.965759 & 5907.755279 & 144.7648273\\
10000 & 81399.38488 & 82103.40372 & 1085.736205\\
100000 & 1048435.866 & 1051292.546 & 8685.889638
\end{array}
$$
