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.

$$\frac{d}{dx}[F_1(x) - F_2(x)] = 0$$Recall, we have previously proven (with the help of the mean value theorem) that functions with a zero derivative must be constant functions.

As such, it must be the case that 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 $f(x)$, also called the **indefinite integral** of $f(x)$, by
$$\int f(x) \, dx$$

The act of finding an indefinite integral is known as **integration** and the symbol $\int$ is known as the **integral sign**.

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 integration rules. For example, one can quickly establish:

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

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

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

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

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

Indeed, all of basic differentiation rules for specific functions (e.g. trigonometric functions, exponential functions, logarithmic functions, etc.) can be reversed, as the results below suggest:

$\displaystyle{\int (\cos x) \, dx = \sin x + C}$

$\displaystyle{\int (\sin x) \, dx = -\cos x + C}$

$\displaystyle{\int \frac{1}{\sqrt{1-x^2}} dx = \arcsin x + C}$

$\displaystyle{\int (a^x \ln a) \, dx = a^x + C}$

$\displaystyle{\int (\sec x \tan x) \, dx = \sec x + C}$

$\displaystyle{\int (\csc x \cot x) \, dx = -\csc x + C}$

$\displaystyle{\int \frac{1}{1+x^2} dx = \arctan x + C}$

$\displaystyle{\int \frac{1}{x \ln a} \, dx = \log_a x + C}$

$\displaystyle{\int (\sec^2 x) \, dx = \tan x + C}$

$\displaystyle{\int (\csc^2 x) \, dx = -\cot x + C}$

$\displaystyle{\int e^x \, dx = e^x + C}$

One may notice none of the above basic integration rules are a reversal of the product, quotient, or chain rules.

Unfortunately, while there is a result connected to reversing the product rule for derivatives that will help us find integrals of some products, and there are means for dealing with integrals of quotients as well -- these techniques of integration will require a little more development. As such, we save them for a future topic of discussion.

That said, there is a way for us to use the chain rule in reverse to help us find the integral of a function, but we save that for the next section.