Too often it's useful to just set $C$ to $0$ after integrating, so I was wondering if there is a symbol for that?
Formally this would mean a shorter way of writing: $\int_{t_0}^x f(t)\,\mathrm{d}t = F(x), (F(t_0) = 0)$
This is useful when integrating sums of integral and splitting these as in: $\int x^2+2\,\mathrm{d}x$.
$(1)\quad \int x^2+2\,\mathrm{d}x = \color{green}{\int_{t_0}^x t^2\,\mathrm{d}t}+\color{red}{\int2\,\mathrm{d}x} = \color{green}{\dfrac{x^3}{3}} + \color{red}{2x + C}$
But this changes the variable from $x$ to $t$, which isn't very nice. Another way is:
$(2)\quad\int x^2+2\,\mathrm{d}x = \color{green}{\int x^2\,\mathrm{d}x}+\color{red}{\int 2\,\mathrm{d}x} = \color{green}{\dfrac{x^3}{3} + C_1} + \color{red}{2x + C_2} = \color{green}{\dfrac{x^3}{3}} + \color{red}{2x} + \color{blue}{C}, \,(\color{green}{C_1}+\color{red}{C_2} = \color{blue}{C})$
Which is sometimes just written as:
$(3)\quad\int x^2+2\,\mathrm{d}x = \color{green}{\int x^2\,\mathrm{d}x}+\color{red}{\int 2\,\mathrm{d}x} = \color{green}{\dfrac{x^3}{3}} + \color{red}{2x} + \color{blue}{C}$
But which I find is missing a step, as it doesn't explain where the $\color{blue}{C}$ comes from. If I could just choose an antiderivative from the first integral $\left(\color{green}{\int x^2\,\mathrm{d}x}\right)$ without a constant, this would be fine.
I was thinking maybe
$(4)\quad \int_{t_0}^x f(t)\,\mathrm{d}t = F(x), (F(t_0) = 0) \overset{_\text{def}}{=} \int_0 f(x)\,\mathrm{d}x = F(x)$
(With the property that $\int_0 f(x)\,\mathrm{d}x + C = \int f(x)\,\mathrm{d}x$)
$(5)\quad \int x^2+2\,\mathrm{d}x = \color{green}{\int_0 x^2\,\mathrm{d}x}+\color{red}{\int x^2+2\,\mathrm{d}x} = \color{green}{\dfrac{x^3}{3}} + \color{red}{2x + C}$
Or in general:
$(6)\quad \int f_1(x)+f_2(x)+\cdots+f_n(x)\,\mathrm{d}x = \int_0 f_1(x)\,\mathrm{d}x+\int_0 f_2(x)\,\mathrm{d}x+\cdots\int f_n(x)\,\mathrm{d}x$
So is there perhaps a better way to write this in an unambiguous way, if not: would my own notation be confused anywhere? (In any case it'd be used only for personal use).
Thanks in advance !