Quite occasionally we see in number theory the below integral evaluated as such:
$$\int_{2}^{x} \frac{dt}{\log t}= \int_{2}^{\sqrt x}\frac{dt}{\log t} + \int_{\sqrt x}^{x}\frac{dt}{\log t} $$
Why the choice of $\sqrt x$ And how does it help facilitate the integral and estimating it.
The first part fits inside a box of width $\sqrt x$ and height $1/\ln2$ so its integral is less than $\sqrt x/\ln 2$.
The second part fits inside a box of width $x$ and height $1/\ln\sqrt x$ so its integral is less than $2x/\ln x$. The first part is narrow and tall, the second part is wide but short. Their combined area is much less than the simplest estimate of $x/\ln 2$ that comes from a single box of width $x$ and height $1/\ln2$.
$\sqrt x$ was chosen to be a fairly simple compromise, where $\sqrt x=o(x)$ and $1/\ln\sqrt x=O(1/\ln x)=o(1/\ln2)$
$$\int_2^{\sqrt x}\frac{dt}{\ln t}\lt\int_2^{\sqrt x}\frac{dt}{\ln2}\\=\frac{\sqrt x -2}{\ln2}$$
$$\int_{\sqrt x}^x\frac{dt}{\ln t}\lt\int_{\sqrt x}^x\frac{dt}{\ln\sqrt x}\\ \lt\frac x{\ln\sqrt x}$$