If $\frac{d}{\mathrm{dx}}F\left(x\right)\=f\left(x\right)$, then an antiderivative of $f\left(x\right)$ is $F\left(x\right)$. In other words, $F\left(x\right)$ is the function whose derivative is $f\left(x\right)$. Thus, given the derivative $f\left(x\right)$, the search for an antiderivative is the search for the function $F\left(x\right)$ for which the derivative is $f\left(x\right)$.

Since the derivatives of $F\left(x\right)$ and $F\left(x\right)\+c$, where $c$ is any constant, are the same function $f\left(x\right)$, the search for an antiderivative of $f\left(x\right)$ does not have a unique answer. There are many different antiderivatives of $f\left(x\right)$, but each differs from the other by an additive constant.

If any one antiderivative is sought, the search might be posed as "Find an antiderivative of $f\left(x\right)$," in which case any single antiderivative would suffice. Alternatively, if the complete family of all possible antiderivatives is sought, the search should be posed as "Find the (most general) antiderivative of $f\left(x\right)\,\"$ in which case any one antiderivative plus an arbitrary additive constant constitutes the complete family of all possible antiderivatives.

There are calculus instructors who will interpret every request for an antiderivative as a request for the complete family, and will therefore insist that antiderivatives must always appear as $F\left(x\right)\+c$. On the other hand, there are calculus instructors who recognize the difference between a single member of the complete family and the complete family itself.

Table 3.10.1 lists an antiderivative for some common functions met while studying differentiation in Chapter 2. The table was constructed by gleaning from lists of derivatives, the functions that were differentiated. Note that neither this table, nor any of the common commercial tables, includes the additive constant. That is left for the user to include (or exclude).$$

The antiderivative of a constant $a$ is $a x$, even if $a\=0$, in which case the antiderivative is rightly given by $0\+c\=c$. In other words, the antiderivative of zero is any constant.

The antiderivative of $x^a$ follows the "Power rule" provided $a\ne -1$. In words, the Power rule is "Add one to the power and divide by the new power." If $a\=-1$, then the antiderivative is the logarithm. The typical calculus text will use absolute values, as in Table 3.10.1, but Maple's antiderivative for $x^{-1}$ is just $\ln \left(x\right)$, a result that is correct on the complex plane.

An antiderivative of $1\/\sqrt{1-x^2}$ can be given as either $\arcsin \left(x\right)$ or $-\arccos \left(x\right)$. However, it should not be inferred from this that $\arcsin \left(x\right)\=-\arccos \left(x\right)$. In fact, the correct identity is $\arcsin \left(x\right)\=\mathrm{\π}\/2-\arccos \left(x\right)$. There are two forms for the antiderivative because the right additive constant transforms one antiderivative into the other.

The typical antiderivative tabulated for $1\/\left(1\+x^2\right)$ is $\arctan \left(x\right)$. The identity $\arctan \left(x\right)\=\mathrm{\π}\/2-\mathrm{arccot}\left(x\right)$ and the additive constant argument accounts for the two different forms of the antiderivative.

It is surprising that the derivatives of $\mathrm{arctanh}\left(x\right)$ and $\mathrm{arccoth}\left(x\right)$ are the same. However, Figure 3.10.1 shows that the domains for these two antiderivatives are different.

Finally, it is worth remembering that an antiderivative for $e^x$ is $e^x$, that is, the exponential function is "its own derivative and its own antiderivative."