Show that $\int_{2}^x\frac{\pi(t)}{t(t-1)}dt=\log \log x+ O(1)$
Do you use the fact that $\pi(t) = \frac{t}{\log t} + O\left(\frac{t}{\log^2t}\right)$ and then
$\int_{2}^x\frac{\pi(t)}{t(t-1)}dt= \int_2^x\left(\frac{t}{\log t} + O\left(\frac{t}{\log^2t}\right)\right)\left(\frac{1}{t(t-1)}\right)dt$
and work from there?
Since $\frac{1}{t-1}=\frac{1}{t}+O\left(\frac{1}{t^2}\right)$ and $\frac{1}{t^2\log t}$, as well as $\frac{1}{t(t-1)\log t}$, is integrable over $[2,+\infty)$, we have: $$\int_{2}^{x}\frac{dt}{(t-1)\log t}=\int_{2}^{x}\frac{dt}{t\log t}+O(1) = \log\log x + O(1).$$