Let us define $r(x)$ by
$$
\sum_{p \leq x} \log p=x(1+r(x))
$$
so we have $r(x) \rightarrow 0$ for $x \rightarrow \infty$ by VII.4.4.
Trivially, the following holds:
$$
\sum_{p \leq x} \log p \leq \pi(x) \log x
$$
and hence
$$
\pi(x) \geq \frac{x}{\log x}(1+r(x))
$$
$$
\pi(x) \leq \frac{x}{\log x}(1+r(x)) q^{-1}+x^{q} .
$$
This inequality can now be specified for a suitable value of $q$, namely for $q=1-$ $1 / \sqrt{\log x}(x \geq 2)$. Then we have
$$
\pi(x) \leq \frac{x}{\log x}(1+R(x))
$$
with
$$
R(x)=-1+(1+r(x))\left(1-\frac{1}{\sqrt{\log x}}\right)^{-1}+(\log x) x^{-1 / \sqrt{\log x}} .
$$
It obviously holds $R(x) \rightarrow 0$ for $x \rightarrow \infty$.
MATH3977 COURSE NOTES ：
For $t \in \mathbb{R}$ we define
$$
\beta(t):=t-[t]-1 / 2 \quad([t]:=\max {n \in \mathbb{Z}, n \leq t})
$$
Then we have $\beta(t+1)=\beta(t)$ and $|\beta(t)| \leq \frac{1}{2}$.
The integral
$$
F(s):=\int_{1}^{\infty} t^{-s-1} \beta(t) d t
$$
converges absolutely for $R e(s)>0$, and represents in this right half-plane an analytic function $F$. For $\operatorname{Re}(s)>1$ it holds:
$$
\zeta(s)=\frac{1}{2}+\frac{1}{s-1}-s F(s)
$$