Can an indefinite integral depend on $c$? If we have an indefinite integral $g(x)+c$ of a function $f,$ can we treat $c$ (entirely independent of $x$) as being some changing value?
For example, if we want to use the parameter $c$ to classify the set of antiderivatives of $f,$ can we define $h_c(x)=g(x)+c=j(x,c)$ under the understanding that $h_c(x)$ represents the value of a different function for different $c$?


If we have an indefinite integral $g(x)+c$ of a function $f,$ can we treat $c$ as being some changing value?
can we define $h_c(x)=g(x)+c$ under the understanding that $h_c(x)$ represents the value of a different function for different $c$?

An indefinite integral is more a useful notational shorthand than a mathematically important object. It is conventionally called a family of antiderivatives; as such, if its integrand is of the form $g'$ and has an interval domain, then, literally, $$\int g'\,\mathrm dx=\{g(x)+C\mid C\in\mathbb R\},$$ where $C$ represents uncountably many values. How to understand the "indefinite integral" notation in calculus? contains several justifications for manipulating and adding indefinite integrals as actual sets by reading $\text‘=\text’$ as an equivalence relation such that writing \begin{align}x^2+C=x^2+3+C\quad&\implies0=3\\x^2+C=x^2+3+D\quad&\implies E=C-D=3\end{align} make sense.
More simply, we can frame an indefinite integral as the general representation of its integrand's antiderivatives, its specification containing one independent parameter (arbitrary constant) $C_i$ per maximal interval of its integrand's domain. Here, each instantiation of $$\int g'\,\mathrm dx=g(x)+C$$ has an unimportant value of $C.$ Even though this object is not a particular antiderivative, we manipulate it as if $C$ is merely undetermined. This seems to match your above description, and hopefully answers your question.
In the context of solving differential equations with given conditions, $C$ becomes an unknown whose value is to be determined.
