Why does every indefinite integral have a constant C?

Because the derivative of any constant is 0, if F(x) is one antiderivative then F(x) + C is also an antiderivative for every constant C. The whole family of antiderivatives is written ∫ f(x) dx = F(x) + C.

What is the formula for the integral of a composite function f(ax+b)?

If ∫ f(x) dx = F(x) + C, then ∫ f(ax+b) dx = 1/aF(ax+b) + C for a ≠ 0. The key is to multiply by the extra factor 1/a out front.

What is the antiderivative of 1/x?

The antiderivative of 1/x is ln|x| + C. The absolute value is required because 1/x is also defined when x < 0.

Table of Common Integrals (Antiderivatives) with Examples

Q: What is an antiderivative?

A function F(x) is an antiderivative of f(x) on an interval K if F'(x) = f(x) for every x in K. In other words, finding an antiderivative is the reverse operation of differentiation.

Integration (finding antiderivatives) is the foundation of every integral you'll compute in calculus. This guide gathers a complete table of common integrals — the basic power, trigonometric, exponential, and logarithmic forms — along with the key properties and the essential rule for handling composite functions of the form $(a x + b)$ , all with detailed worked examples.

What is an antiderivative?

Let $f (x)$ be a function defined on an interval $K$ . A function $F (x)$ is called an antiderivative of $f (x)$ on $K$ if:

$F^{'} (x) = f (x) for all x \in K$

Because the derivative of a constant is $0$ , if $F (x)$ is one antiderivative then $F (x) + C$ (where $C$ is any constant) is also an antiderivative. So the entire family of antiderivatives of $f (x)$ is written with the indefinite integral sign:

$\int f (x) d x = F (x) + C$

Never forget the constant $C$ — it's the single most common point deduction on exams.

Basic integral table

The formulas below are the ones to memorize. You can verify each one by differentiating the right-hand side to recover the integrand.

Function $f (x)$	Integral $\int f (x) d x$
$0$	$C$
$k$ (constant)	$k x + C$
$x^{α} (α \neq = - 1)$	$\frac{x ^{α + 1}}{α + 1} + C$
$\frac{1}{x}$	$ln ∣ x ∣ + C$
$\frac{1}{x ^{2}}$	$- \frac{1}{x} + C$
$\frac{1}{x}$	$2 x + C$
$e^{x}$	$e^{x} + C$
$a^{x} (0 < a \neq = 1)$	$\frac{a ^{x}}{ln a} + C$

Integrals of trigonometric functions

Function $f (x)$	Integral $\int f (x) d x$
$cos x$	$sin x + C$
$sin x$	$- cos x + C$
$\frac{1}{cos ^{2} x}$	$tan x + C$
$\frac{1}{sin ^{2} x}$	$- cot x + C$

Watch the sign pair that trips everyone up: the integral of $cos x$ is $+ sin x$ , but the integral of $sin x$ is $- cos x$ .

Properties of antiderivatives

Antiderivatives are linear, which means you can split sums apart and pull constants outside the integral sign:

$\int k \cdot f (x) d x = k \int f (x) d x (k \neq = 0)$

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

These two properties let you break a complicated expression into several basic integrals and add the results back together. There's also the fundamental link back to differentiation:

$(\int f (x) d x)^{'} = f (x)$

Composite functions of the form (ax + b)

This is an extremely handy extension. If $\int f (x) d x = F (x) + C$ , then for $a \neq = 0$ :

$\int f (a x + b) d x = \frac{1}{a} F (a x + b) + C$

The crucial step: multiply by the extra factor $\frac{1}{a}$ out front. Here are the forms you'll meet most often:

Function	Integral
$(a x + b)^{α} (α \neq = - 1)$	$\frac{1}{a} \cdot \frac{( a x + b ) ^{α + 1}}{α + 1} + C$
$\frac{1}{a x + b}$	$\frac{1}{a} ln ∣ a x + b ∣ + C$
$e^{a x + b}$	$\frac{1}{a} e^{a x + b} + C$
$cos (a x + b)$	$\frac{1}{a} sin (a x + b) + C$
$sin (a x + b)$	$- \frac{1}{a} cos (a x + b) + C$

Worked examples

Example 1 — using linearity. Evaluate $\int (3 x^{2} - 4 x + 5) d x$ .

Split the integral term by term:

$\int 3 x^{2} d x - \int 4 x d x + \int 5 d x = 3 \cdot \frac{x ^{3}}{3} - 4 \cdot \frac{x ^{2}}{2} + 5 x + C$

$= x^{3} - 2 x^{2} + 5 x + C$

Example 2 — powers and roots. Evaluate $\int (x + \frac{1}{x ^{2}}) d x$ .

Rewrite $x = x^{1/2}$ and $\frac{1}{x ^{2}} = x^{- 2}$ , then apply the power rule for $x^{α}$ :

$\int x^{1/2} d x + \int x^{- 2} d x = \frac{x ^{3/2}}{3/2} + \frac{x ^{- 1}}{- 1} + C = \frac{2}{3} x x - \frac{1}{x} + C$

Example 3 — a composite $(a x + b)$ . Evaluate $\int e^{2 x + 1} d x$ .

Here $a = 2$ , so multiply by $\frac{1}{2}$ :

$\int e^{2 x + 1} d x = \frac{1}{2} e^{2 x + 1} + C$

Example 4 — a composite trig function. Evaluate $\int cos (3 x - 2) d x$ .

$\int cos (3 x - 2) d x = \frac{1}{3} sin (3 x - 2) + C$

Common mistakes

Forgetting the constant $C$ : every indefinite integral needs $+ C$ ; leaving it off costs points.
Dropping the $\frac{1}{a}$ factor in the $(a x + b)$ forms — this is the most frequent error when integrating $e^{a x + b}$ or $sin (a x + b)$ .
Getting the trig sign wrong: remember $\int sin x d x = - cos x + C$ (with a minus sign), while $\int cos x d x = sin x + C$ .
Omitting the absolute value in $\int \frac{1}{x} d x = ln ∣ x ∣ + C$ — the bars are required because the function is defined for $x < 0$ too.
Misusing the power rule when $α = - 1$ : the case $\frac{1}{x}$ does not use $\frac{x ^{α + 1}}{α + 1}$ ; it gives $ln ∣ x ∣$ instead.

Once you've memorized the table and remembered to multiply by $\frac{1}{a}$ for composite forms, you can handle almost every basic integral you'll meet. Try the examples above in the calculator to check each step of your solution.