Is indefinite integration non-linear? Let us consider this small problem:
$$
\int0\;dx = 0\cdot\int1\;dx = 0\cdot(x+c) = 0
\tag1
$$
$$
\frac{dc}{dx} = 0 \qquad\iff\qquad
\int 0\;dx = c, \qquad\forall c\in\mathbb{R}
\tag2
$$
These are two conflicting results. Based on this other question, Sam Dehority's answer seems to indicate:
$$
\int\alpha f(x)\;dx\neq\alpha\int f(x)\;dx,\qquad\forall\alpha\in\mathbb{R}
\tag3
$$
However, this clearly implies that indefinite integration is nonlinear, since a linear operator $P$ must satisfy $P(\alpha f) = \alpha Pf, \forall\alpha\in\mathbb{R}$, including $\alpha=0$. After all, a linear combination of elements of a vector space $V$ may have zero valued scalars: $f = \alpha g + \beta h, \forall\alpha,\beta\in\mathbb{R}$ and $g, h\in V$. This all seems to corroborate that zero is not excluded when it comes to possible scalars of linear operators.
To take two examples, both matrix operators in linear algebra and derivative operators are linear, even when the scalar is zero. In a matrix case for instance, let the operator $A$ operate a vector: $A\vec{x} = \vec{y}$. Now: $A(\alpha\vec{x}) = \alpha A\vec{x} = \alpha\vec{y}$. This works even for $\alpha = 0$.
Why is $(3)$ true? Can someone prove it formally? If $(3)$ is false, how do we fix $(1)$ and $(2)$? When exactly does the following equality hold (formal proof)?
$$
\int\alpha f(x)\;dx  = \alpha\int f(x)\;dx,\qquad\forall\alpha\in\mathbb{R}
$$
I would appreciate formal answers and proofs. 

My "answer" should be a comment under the two posted answers, but I don't have enough reputation to post comments. There is a (very popular) mistake in both answers. Indefinite integral operator does NOT give a class of functions equal up to constant translation. Let's look at an example. Someone might evaluate integral of 1/x like this:
$$
\int\frac{1}{x}dx = 
\begin{cases}
      logx + C & \text{if x > 0}\\
      log(-x) + C & \text{if x < 0}\\
    \end{cases}
= log|x| + C   \\
\text{where C} \in \mathbb{R}.
$$
However, this is not an exhaustive answer. The constants for negative and positive x can differ. The exhaustive answer would be
$$
\int\frac{1}{x}dx = 
\begin{cases}
      logx + C & \text{if x > 0}\\
      log(-x) + D & \text{if x < 0}\\
    \end{cases},\\
\text{where C,D} \in \mathbb{R}.
$$
Thus it is not true to say that
$$
\int  f(x) dx = \{ F(x) + C : C \in \mathbb{R} \}\\
\text{for a particular $F(x)$}.
$$
