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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.