Avoiding the Integration Constant

I don't think that there is a standard way of doing it but of course, you can invent your on. In this case, an equivalence relation or rather a sequence of equivalence relations:

For functions $f, g$, let $f \equiv_d g$ if there exists a polynomial $p$ of degree at most $d$, such that $f = g +p$. In this case, you want $\deg 0 = -1$.

We have $f \equiv_{-1} g$ iff $f=g$ and integrating raises the degree $d$ by one.

Your example becomes $$\int f'(x) \operatorname{d}x \equiv_0 f(x) = f_1(x) = \dots = f_n(x)$$ and in a second step $$\int f'(x) \operatorname{d}x = f_n(x) + C.$$

Edit: In this answer, I defined the symbol $\equiv$ (if you don't like it, you can replace it by something else, say $\sim$). Another way to phrase the definition is the following: For functions $f, g$ and $d \in \{-1, 0, 1, 2, \dots \}$, let $f \equiv_d g$ if the difference $f - g$ is a polynomial of degree at most $d$. For every $d$, $\equiv_d$ is an equivalence relation – the proof is easy.

Slightly abusing notation (I will identify a function $f$ with its term $f(x)$ for brevity), we have for example

• $f(x) \equiv_{-1} g(x)$ if and only if $f = g$, because the only degree-$(-1)$ polynomial is the zero polynomial, thus $f - g = 0$, i.e. $f = g$.
• $\sin x \equiv_0 \sin x + 15$, because $15$ is a degree-$0$ polynomial
• $f(x) \equiv_d f(x) + c_d x^d + c_{d-1} x^{d-1} + \dots + c_0$, because $c_d x^d + c_{d-1} x^{d-1} + \dots + c_0$ is a degree-$d$ polynomial

Now, whenever $f''(x) = g''(x)$, we know that $f'(x) + C = g'(x) + D$ with constants $C, D \in \mathbb{R}$. Thus, $f'(x) - g'(x) = D - C$ which is constant, i.e. a degree-$0$ polynomial. Using the notation defined above, we can write $f \equiv_0 g$, eliminating the explicit constant.

This also works for higher degrees. Say we know $f'(x) \equiv_d g'(x)$. Then there exists a degree-$d$ polynomial $p$ or more explicitly real numbers $c_0, \dots, c_d$, such that $f'(x) - g'(x) = p(x) = c_d x^d + \dots + c_0$. Integration: $$f(x) - g(x) = \frac{c_d}{d+1} x^{d+1} + \dots + c_0 x + C$$ which is a degree-$(d+1)$ polynomial. Thus $f \equiv_{d+1} g$.

The nice thing about an equivalence relation is that it behaves in many ways like equality. For example, you can use it in an equality chain like $$f_0(x) = f_1(x) \equiv_d f_2(x) = f_3(x)$$ if you keep in mind that you can only deduce $f_0(x) \equiv_d f_3(x)$ from that and not $f_0(x) = f_3(0)$.

Also note that this notation is not standard. If you want to use it and others to understand it, you will have to give the definition.

You could reduce the number of your constants by half by using only one constant each time you integrate. For example, when solving $$r'(x)=s'(x)$$ write the answer as $$r(x)=s(x)+C_1$$ rather than (as you were doing) $$r(x)+C_1=s(x)+C_2$$ Using only one constant when integrating both sides of an equation is perfectly rigorous and the proof is obvious.

ADDED:

1) Another possibility is to use constants like $c,d,e,\ldots$ (or $k,l,m,\ldots$ or $p,q,r,\ldots$ or $\alpha,\beta,\gamma,\ldots$) rather than $C_1,C_2,C_3,\ldots$. This does not reduce the number of constants but it does reduce the tiresome subscripts and make it look like there are fewer constants.

2) When you wrote "I often just combine as many constants in one shot without mercy..." I thought you meant things like changing $\pm C_1e^{x+C_2}$ to $C_1e^x$, which is less rigorous. Using only one constant per integration is a no-brainer.

Here is one way around your problem:

$\int f'(x) \mathrm dx = f(x) + C$ for some $C$. So

$$\begin{align} \int f'(x)\ \mathrm dx - C &= f(x) \\ &= f_1(x) \\ &= f_2 (x) \\ &= f_3 (x) \\ &= \dots \\ &= f_n(x) \\ \end{align}$$

Therefore $\int f'(x) \mathrm dx = f_n(x) + C$ for some $C$.

I don't think this is exactly what you were looking for, but then I suspect that what you are looking for doesn't exist.

For the easy situation of one integration, you can write

$$ ∫f(x)dx ∋ F(x) = F_1(x) = … = F_n(x)\\ \text{hence} ∫f(x)dx = F_n(x) + C $$

It's because you can say that primitive operator function really gives you the set of all primitive functions, all of which differs by a constant. Then, to be consistent, you can say that the second equality is a set equality and $C$ is a set of all constants.