Source: P36 of Elementary Differential Equations, 9th Ed by Boyce, DiPrima et al.
${\int{f(t)\text{ }dt} = \int_{t_0}^t f(s) \text{ } ds \quad \text{ where $t_0$ is some convenient lower limit of integration.}} \tag{$*$}$
$\Large{\text{1.}}$ I now know: $ \int{f(t)\text{ }dt} \qquad$ is $\color{green}{\text{ a set of functions $\qquad$ (**)}} $
and $ \int_{t_0}^t f(s) \text{ } ds \qquad$ is $\color{#B53389}{\text{ an element of set (**) above.}} \quad $
So how and why is $(*)$ true? How can a $\color{green}{\text{ a set of functions}}$ = $\color{#B53389}{\text{ an element of the same set}} $ ?
Supplementary to William Stagner's answer and Peter Tamaroff's comment
Thanks to your explanations, I now know that: $\int{f(t)}\text{ }dt = g(t) + C \qquad \forall \ C \in \mathbb{R}\ \tag{$\natural$}$ $\int_{t_0}^t f(s) \text{ } ds = g(t) - g(t_0) \tag{$\blacklozenge$}$
Since $g(t)$ is one function and $t_0$ is one arbitrarily chosen argument/number,
thus $-g\left(t_0\right)$ is ONE FIXED number.
In contrast, $C$ is ANY real number.
$\Large{\text{3.}}$ So $(\natural) \mathop{=}^{?} \, (\blacklozenge) \iff C \mathop{=}^{?} -g(t_0).$ But how and why is : $ C \mathop{=}^{?} -g\left(t_0\right) \; $?