antiderivative
Let $I$ be an open interval of $\mathbb{R}$ and $f:I\u27f6\mathbb{R}$ a real function.
A function $F:I\u27f6\mathbb{R}$ is called an antiderivative or a primitive of $f$ if $F$ is differentiable^{} and its derivative^{} is equal to $f$, i.e.
$${F}^{\prime}(x)=f(x)\mathit{\hspace{1em}}\text{for all}x\in I.$$ 
Note that there are an infinite^{} number of antiderivatives for any function $f$ since any constant can be added or subtracted from any valid antiderivative to yield another equally valid antiderivative.
To account for this, we express the general antiderivative, or indefinite integral, as follows:
$$\int f(x)\mathit{d}x=F+C$$ 
where $C$ is an arbitrary constant called the constant of integration. The $dx$ portion means “with respect to $x$”, because after all, our functions $F$ and $f$ are functions of $x$.
There is no loss in generality with this notation since in fact all antiderivatives of $f$ take this form as the following theorem demonstrates:
Theorem. Let $F\mathrm{,}G$ be two antiderivatives of a given function $f$ defined on an open interval $I$. Then $F\mathrm{}G\mathrm{=}\text{\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}}$.
Proof. Since ${F}^{\prime}(x)=f(x)$ and ${G}^{\prime}(x)=f(x)$, we have ${F}^{\prime}(x){G}^{\prime}(x)=0$ on the whole $I$. Thus, by the fundamental theorem of integral calculus, $F(x)G(x)=\text{const}$. $\mathrm{\square}$
This is no longer true if the domain of the function $f$ is not an open interval (is not connected^{}). For that scenario, the following more general result holds:
Theorem. Let $U\mathrm{\subset}\mathrm{R}$ be an open set (not necessarily an interval). Suppose $F\mathrm{,}G$ are antiderivatives of a given function $f\mathrm{:}U\mathrm{\u27f6}\mathrm{R}$. Then $F\mathrm{}G$ is constant in each connected component^{} of $U$ (each interval in $U$).
For example, consider the function $f:\mathbb{R}\setminus \{0\}\u27f6\mathbb{R}$ given by $f(x)=\frac{1}{x}$. Notice that the domain of $f$ is not an interval, but the union of the disjoint intervals $(\mathrm{\infty},\mathrm{\hspace{0.17em}0})$ and $(0,+\mathrm{\infty})$. Then, all the antiderivatives of $f$ take the form
$$ 
0.1 Remarks

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For complex functions, the definition of antiderivative is exactly the same and the above results also hold (one just needs to consider “connected open subsets” instead of “open intervals”).
Title  antiderivative 

Canonical name  Antiderivative 
Date of creation  20130322 12:14:55 
Last modified on  20130322 12:14:55 
Owner  asteroid (17536) 
Last modified by  asteroid (17536) 
Numerical id  21 
Author  asteroid (17536) 
Entry type  Definition 
Classification  msc 26A36 
Synonym  general antiderivative 
Synonym  indefinite integral 
Synonym  primitive 
Related topic  AntiderivativeOfComplexFunction 
Defines  constant of integration 