Removing the $+c$ from the antiderivative While I was first learning about integrals and antiderivatives, I took for granted that, for example,  $\int 2xdx = x^2 + c$. But if the constant can be any number, that means that $\int 2xdx = x^2 = x^2 + 1$ which doesn't make any sense. I understand that the constant is usually just left as $c$, but it is implied that it can be any number, right? We don't define the square root of $4$ to be $2$ and $-2$ and we don't define $\arcsin(0)$ to be $0$, $2\pi$, $4\pi$...
If you try to define $\int\frac{d}{dx}(f(x))dx = f(x)$, a problem arises. Mainly, if $f(x)$ is something like $x^3 + 4$. The $4$ gets lost when taking the derivative so we can't be sure of what a function is if we only know its derivative.
A potential solution to this problem would be to define a function $S(f)$ such that it takes a function as its input and outputs a function stripped of its constant e.g. $S(x^3 + 4) = x^3$. If this is done, the antiderivative can be defined as follows: $$\int \frac{d}{dx}(f(x))dx = S(f(x))$$ $$\frac{d}{dx}(\int f(x)dx)=f(x)$$
Is it possible to define $S(f)$? If not, is there any other way to define the antiderivative so that it doesn't have a constant?
 A: The integral, while commonly thought of as an inverse to the differentiation operator, is not really a "true" inverse.
Just like ${f(x)=x^2}$ is, without any extra constraints, technically is not invertible. It lacks injectivity. ${f(x)=f(y)}$ does not imply that ${x=y}$. If you want an inverse - you must either resort to multi-valued functions, or you must restrict the domain you are interested in. If we do the latter, then indeed
$${f^{-1}(x) = \sqrt{x}}$$
In exactly the same way,
$${\frac{df}{dx}=\frac{dg}{dx}}$$
does not imply ${f(x)=g(x)}$. The differential operator clearly lacks injectivity, and you gave concrete examples
$${\frac{d}{dx}\left(x^2 + 1\right)=\frac{d}{dx}\left(x^2 \right)}$$
but obviously ${x^2 + 1\neq x^2\ \forall\ x \in \mathbb{R}}$.
Making the argument that you can set ${c}$ to be whatever you want in two different contexts and claiming the results are still equal is an invalid step. It's like saying ${2^2 = (-2)^2 = 4}$, so ${2=-2}$ - injectivity does not hold.
Now, when it comes to evaluating definite integrals - it doesn't matter which anti-derivative you pick. The Fundamental Theorem of Calculus simply states that if ${f(x)}$ is continuous on ${[a,b]}$ then
$${\int_{a}^{b}f(x)dx=F(b)-F(a)}$$
where ${F(x)}$ is any function satisfying ${\frac{d}{dx}\left(F(x)\right)=f(x)}$. So it doesn't matter what constant you specify. And it's easy to see why:
$${\left(F(b) + c\right) - \left(F(a) + c\right)=F(b)-F(a)}$$
To summarise: you cannot do anything about the missing information (unless in context you can Algebraically find out the value of the ${+c}$ constant) and it's not a bug. It's a feature! (Sorry for the terrible overused Computer Science reference).
