From the prime number theorem we know that $\pi(x)\sim x/\log x$, i.e. $\dfrac{\pi(x)\log x}{x}\rightarrow 1$ as $x\rightarrow \infty$.
How can we use that to show that $\pi(x)\sim\int_2^x\dfrac{1}{\log t}dt$? The integral $\dfrac{1}{\log t}$ does not have a closed form, as far as I know.
You actually have it backwards. The Prime Number Theorem shows that $\pi(x) \sim \int_2^x \frac{1}{\log t} dt,$ which is a much better approximation than $\pi(x) \sim \frac{x}{\log x}.$ To get the "standard form" from the first one, just integrate by parts.
To answer Daniel Fischer's comment, see this part of the wiki article on the PNT.