Antiderivatives and indefinite integrals | Calculus

Running derivatives backward

Differentiation turns a function into its rate of change. Now we go the other way: given a derivative, find the function.

$$x^3$$ is an antiderivative of $$3x^2$$ . So are $$x^3 + 7$$ and $$x^3 - 2$$ , since constants have derivative $$0$$ . All antiderivatives of a function differ by a constant.

Basic rules

RuleBasic integration rules

$$f(x)$$	$\int f(x)\,dx$
$$x^n$$ , $n \neq -1$	$\frac{x^{n + 1}}{n + 1} + C$
$\frac{1}{x}$	$\ln\lvert x\rvert + C$
$$e^x$$	$$e^x + C$$
$\sin x$	$-\cos x + C$
$\cos x$	$\sin x + C$
$\sec^2 x$	$\tan x + C$

Constants factor out, and the integral of a sum is the sum of the integrals.

You can always check an answer by differentiating it.

U-substitution

U-substitution reverses the chain rule. Look for an "inside" function $$u$$ whose derivative also appears in the integrand.

Initial value problems

An extra condition picks out a single antiderivative.

Practice

Quick check

What is $\int x^6\,dx$ ?

Quick check

For $\int 3x^2(x^3 + 5)^4\,dx$ , which substitution works best?

$\(f(x)\)$	$\int f(x)\,dx$
$\(x^n\)$ , $n \neq -1$	$\frac{x^{n + 1}}{n + 1} + C$
$\frac{1}{x}$	$\ln\lvert x\rvert + C$
$\(e^x\)$	$\(e^x + C\)$
$\sin x$	$-\cos x + C$
$\cos x$	$\sin x + C$
$\sec^2 x$	$\tan x + C$