For $\frac{d}{dx}\text{arcsec}(x)=\frac{1}{x\sqrt{x^2-1}}$ and $\int{\frac{1}{x\sqrt{x^2-1}}dx}=\text{arcsec}(x)$, shouldn't the condition be $|x|>1$?

$$\int{\frac{1}{\sqrt{1-x^2}}dx=\arcsin(x)}\tag{4},\quad |x|<1$$ $$\frac{d}{dx}{\arcsin(x)}=\frac{1}{\sqrt{1-x^2}}\tag{3}$$

1. For $(3)$, the condition should be $|x|<1$, shouldn't it?

Notice that

• $\displaystyle\frac{1}{\sqrt{1-x^2}}$ is defined on only $(-1,1);$
• $\arcsin(x)$ is defined on only $[-1,1];$
• the derivative of $\arcsin(x)$ is undefined when $x=\pm1.$

Thus, your suggested condition is already implicit in $(4)$'s integrand (but no harm explicating it), while in $(3),$ I think that it's good practice to explicitly specify either $x\ne\pm1$ or, as you suggest, $|x|<1.$

$$\frac{d}{dx}\operatorname{arcsec}(x)=\frac{1}{x\sqrt{x^2-1}}\tag{1}$$ $$\int{\frac{1}{x\sqrt{x^2-1}}dx}=\operatorname{arcsec}(x)\tag{2}$$

2. For $(1)$ and $(2)$, the condition should be $|x|>1$, shouldn't it?

$(1)$ is actually incorrect (at least for negative $x$), because in fact, $$\frac{\mathrm d}{\mathrm dx}\operatorname{arcsec}(x)=\frac{1}{\color{red}|x\color{red}|\sqrt{x^2-1}}\quad\left(|x|>1\right).\tag{1c}$$

$(2)$ is also actually incorrect, because in fact, $$\int\frac{\mathrm dx}{x\sqrt{x^2-1}}= \operatorname{arcsec}\color{red}|x\color{red}|+C.\tag{2c}$$

As for your question about the conditions:

$\operatorname{arcsec}(x)$ is defined, but has no derivative, at $1$ and $-1,$ so we exclude these points from its domain in $(1c);$

$(2c)$ doesn't require the condition $|x|>1$ explicitly since its integrand is defined only there anyway.