Antiderivatives and Indefinite Integrals

Running derivatives backward

An antiderivative of $f$ is any function $F$ with $F'(x) = f(x)$. Since $\tfrac{d}{dx}x^3 = 3x^2$, the function $x^3$ is an antiderivative of $3x^2$. So are $x^3 + 5$ and $x^3 - \pi$: a constant disappears when differentiated.

The indefinite integral collects all of them:

$$\int 3x^2\,dx = x^3 + C$$

where $C$ is an arbitrary constant. By the Fundamental Theorem, antiderivatives are exactly what definite integrals need.

Why + C covers them all

Suppose $F$ and $G$ are both antiderivatives of $f$ on an interval. Then $(G - F)' = f - f = 0$ everywhere on it. The Mean Value Theorem showed that a function with zero derivative on an interval is constant. So $G = F + C$.

Every antiderivative is one particular antiderivative plus a constant, and every constant is possible. The $+ C$ is not decoration; leaving it out claims there is only one answer.

The basic rules

Each row is a derivative rule read from right to left:

Integral	Antiderivative
$\int x^n\,dx$, $n \ne -1$	$\tfrac{x^{n+1}}{n + 1} + C$
$\int \tfrac{1}{x}\,dx$	$\ln\lvert x \rvert + C$
$\int e^x\,dx$	$e^x + C$
$\int a^x\,dx$	$\tfrac{a^x}{\ln a} + C$
$\int \sin x\,dx$	$-\cos x + C$
$\int \cos x\,dx$	$\sin x + C$
$\int \sec^2 x\,dx$	$\tan x + C$
$\int \tfrac{1}{1 + x^2}\,dx$	$\arctan x + C$
$\int \tfrac{1}{\sqrt{1 - x^2}}\,dx$	$\arcsin x + C$

Integrals of sums split into sums, and constant factors come out, exactly as for derivatives. The absolute value in $\ln\lvert x \rvert$ covers negative $x$: there $\tfrac{d}{dx}\ln(-x) = \tfrac{-1}{-x} = \tfrac{1}{x}$.

Initial value problems

An initial value problem gives a derivative and one value of the function. The value fixes $C$, leaving a single answer.

This is how motion is recovered: velocity is an antiderivative of acceleration, and position an antiderivative of velocity. The initial velocity and position fix the two constants.

Worked examples

Common mistakes

Practice problems