A function $F(x)$ is defined to be an *antiderivative* of a function $f(x)$ when $F'(x) = f(x)$.

For a given function, we might have many antiderivatives. Consider $f(x)=3x^2 + \cos x$. We can easily see that the derivatives of $$\begin{array}{ccl} F_1(x) &=& x^3 +\sin x\\\\ F_2(x) &=& x^3 +\sin x + 7\\\\ F_3(x) &=& x^3 +\sin x - \sqrt{5} \end{array}$$ all agree with $f(x)$, and are consequently all antiderivatives of $f(x)$.

The examples above suggest that we can find additional antiderivatives for a given function, once we have one of them, by simply adding a constant on the end. One might wonder if there are any functions that differ by more than a constant that share the same derivative. To answer this question, consider the difference between functions $F_1(x)$ and $F_2(x)$, that are both antiderivatives of some function $f(x)$. More specifically, consider the derivative of this difference: $$\frac{d}{dx}[F_1(x) - F_2(x)] = F_1'(x) - F_2'(x) = f(x) - f(x) = 0$$ So the derivative of this difference is zero. What kind of functions have a derivative of zero? Certainly constant functions behave in this way. Could a non-constant function have a zero derivative?

Interestingly, the Mean Value Theorem can quickly resolve this question. For simplicity sake, let us suppose $g'(x)=0$ for all real values of $x$, but our argument is easily modified to account for cases when we only care about the behavior of our function in some open interval. Now, consider the interval $[x_1,x_2]$. By the Mean Value Theorem, there must be some $c$ in $(x_1,x_2)$ where $$g'(c) = \frac{g(x_2)-g(x_1)}{x_2-x_1}$$ But the derivative of $g(x)$ is zero everywhere, including at $x=c$, so $g'(c)=0$. But then, $g(x_2)-g(x_1)=0$, and consequently $g(x_2)=g(x_1)$. Since our choice of $x_1$ and $x_2$ was arbitrary, $g(x)$ must be a constant function.

Thus, if $g'(x) = 0$, then it must be the case that $g(x) = c$ for some constant $c$.

Going back to our discussion about $F_1(x)$ and $F_2(x)$ -- we observed that $$\frac{d}{dx}[F_1(x) - F_2(x)] = 0$$ Hence, for some constant $c$, $$F_1(x) - F_2(x) = c$$ Or equivalently $$F_2(x) = F_1(x) + c$$ So if we can get our hands on a single antiderivative, $F(x)$, of a given function, $f(x)$, we can describe the whole set of antiderivatives of $f(x)$. This set contains every function of the form $$F(x)+c$$ where $c$ is a constant.

For reasons that become clear once the Fundamental Theorem of Calculus is understood, we will denote the set of all antiderivatives of $f(x)$ by
$$\int f(x) dx$$
but given the above analysis, once we know of a single antiderivative $F(x)$ to the function $f(x)$, we can write:
$$\int f(x) dx = F(x) + C$$
where $C$ is to be thought of as *any* constant.

It is not hard to see that many of the basic differentiation rules can be "worked backwards" to produce some basic antidifferentiation rules. In particular, one might notice:

$\displaystyle{\int dx = x + C}$

$\displaystyle{\int k f(x) dx = k \int f(x) dx}$

$\displaystyle{\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx}$

$\displaystyle{\int x^n dx = \frac{x^{n+1}}{n+1} + C}$, when $n \neq 1$

$\displaystyle{\int \frac{1}{x} dx = \ln |x| + C}$