Antiderivatives and Indefinite Integrals

Definition

An antiderivative of $f(x)$ is a function $F(x)$ such that:

$F'(x) = f(x)$

The indefinite integral is the family of all antiderivatives:

$\int f(x) \, dx = F(x) + C$

where $C$ is the constant of integration.

Notation

Symbol	Meaning
$\int$	integral sign
$f(x)$	integrand
$dx$	variable of integration
$F(x)$	antiderivative
$C$	constant of integration

Why +C?

If $F(x)$ is an antiderivative of $f(x)$, then so is $F(x) + C$ for any constant $C$, because:

$\frac{d}{dx}[F(x) + C] = F'(x) + 0 = f(x)$

Basic Antiderivative Table

$f(x)$	$\int f(x) \, dx$
$0$	$C$
$k$ (constant)	$kx + C$
$x^n$ $(n \neq -1)$	$\frac{x^{n+1}}{n+1} + C$
$\frac{1}{x}$	$\ln
$e^x$	$e^x + C$
$a^x$	$\frac{a^x}{\ln(a)} + C$
$\sin(x)$	$-\cos(x) + C$
$\cos(x)$	$\sin(x) + C$
$\sec^2(x)$	$\tan(x) + C$
$\csc^2(x)$	$-\cot(x) + C$
$\sec(x)\tan(x)$	$\sec(x) + C$
$\csc(x)\cot(x)$	$-\csc(x) + C$
$\frac{1}{\sqrt{1-x^2}}$	$\arcsin(x) + C$
$\frac{1}{1+x^2}$	$\arctan(x) + C$

Properties

Constant Multiple Rule

$\int k \cdot f(x) \, dx = k \cdot \int f(x) \, dx$

Sum/Difference Rule

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

Examples

Example 1: Power rule

$\int x^4 \, dx = \frac{x^5}{5} + C$

Example 2: Negative exponents

$\int \frac{1}{x^3} \, dx = \int x^{-3} \, dx = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C$

Example 3: Fractional exponents

$\int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C$

Example 4: Sum of terms

$\int (3x^2 - 2x + 5) \, dx = x^3 - x^2 + 5x + C$

Example 5: Rewriting before integrating

$\int \frac{(x + 1)^2}{x} \, dx = \int \frac{x^2 + 2x + 1}{x} \, dx$

$= \int \left(x + 2 + \frac{1}{x}\right) dx$

$= \frac{x^2}{2} + 2x + \ln|x| + C$

Example 6: Trigonometric

$\int (\sin x + 3\cos x) \, dx = -\cos x + 3\sin x + C$

Example 7: Exponential

$\int (e^x + 2^x) \, dx = e^x + \frac{2^x}{\ln(2)} + C$

Finding a Particular Solution

Given an initial condition, solve for $C$.

Example

Find $f(x)$ if $f'(x) = 6x^2 - 4x + 1$ and $f(1) = 5$.

$f(x) = \int (6x^2 - 4x + 1) \, dx = 2x^3 - 2x^2 + x + C$

$f(1) = 2 - 2 + 1 + C = 5$

$C = 4$

$f(x) = 2x^3 - 2x^2 + x + 4$