When you differentiate x³, you get 3x². Going backwards — asking "what function differentiates to give 3x²?" — is finding an antiderivative. It is differentiation in reverse, and it is the foundation of integration.

The answer is not unique. x³ works, but so does x³+5 or x³−100, because constants vanish under differentiation. This is why every antiderivative carries that "+C" you see everywhere.

What Is an Antiderivative?

An antiderivative of f(x) is any function F(x) such that F'(x) = f(x). The process of finding antiderivatives is called antidifferentiation or indefinite integration.

If F'(x) = f(x), then F is an antiderivative of f

Note that antiderivatives are not unique — if F is one antiderivative, then F + C is also an antiderivative for any constant C. This is why indefinite integrals always include "+ C".

Notation

∫ f(x) dx = F(x) + C

The ∫ symbol (elongated S for "sum") and dx together indicate antidifferentiation with respect to x.

Basic Antiderivative Rules

∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1) ∫ eˣ dx = eˣ + C ∫ 1/x dx = ln|x| + C ∫ sin(x) dx = −cos(x) + C ∫ cos(x) dx = sin(x) + C
📋 Worked Examples
1∫ x³ dx = x⁴/4 + C
2∫ (3x² + 2x) dx = x³ + x² + C
3∫ e^(2x) dx = e^(2x)/2 + C

Connection to the Definite Integral

The Fundamental Theorem of Calculus connects antiderivatives to definite integrals: ∫ₐᵇ f(x) dx = F(b) − F(a). Every antiderivative technique you learn is therefore also a technique for evaluating definite integrals.

Frequently Asked Questions

Why do we add C?▾
Because every constant has derivative zero, F(x) + C has the same derivative as F(x) for any constant C. The "+ C" represents all possible antiderivatives — an entire family of functions, not just one.
Does every function have an antiderivative?▾
Every continuous function has an antiderivative (this is Part 1 of the FTC). But not every antiderivative can be written in terms of elementary functions. For example, e^(−x²) has an antiderivative, but it cannot be expressed using standard functions.

The Family of Antiderivatives

An antiderivative is never unique. If F is one antiderivative of f, then F + 1, F + 100, and F − π are all antiderivatives too — because differentiating any constant gives zero. This is why we write ∫f(x)dx = F(x) + C. The constant C represents the entire family, and different values of C give different antiderivatives that are all equally valid.

Geometrically: the family F(x) + C is a set of curves that are all vertical shifts of each other. They all have the same slope at each x-value (which equals f(x)), but they pass through different y-intercepts.

Finding C from Initial Conditions

In applications, you often know a specific value of the antiderivative at one point. This pins down C uniquely — turning the family F(x) + C into a single function.

📋 Finding the specific antiderivative
ProblemFind F(x) if F'(x) = 3x² and F(0) = 5
GeneralF(x) = x³ + C
Apply F(0)=55 = 0 + C → C = 5
AnswerF(x) = x³ + 5
📋 Position from acceleration (physics)
Givena(t) = −10 m/s² (gravity), v(0) = 20 m/s, s(0) = 0
Velocityv(t) = ∫a dt = −10t + C₁. v(0)=20 → C₁=20. So v(t) = 20 − 10t
Positions(t) = ∫v dt = 20t − 5t² + C₂. s(0)=0 → C₂=0. So s(t) = 20t − 5t²

Extended Antiderivative Table

∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1) ∫ 1/x dx = ln|x| + C ∫ eˣ dx = eˣ + C ∫ aˣ dx = aˣ/ln(a) + C ∫ sin x dx = −cos x + C ∫ cos x dx = sin x + C ∫ sec²x dx = tan x + C ∫ csc²x dx = −cot x + C ∫ sec x tan x dx = sec x + C ∫ 1/√(1−x²) dx = arcsin x + C ∫ 1/(1+x²) dx = arctan x + C

Always Check Your Answer

Differentiate your answer and verify you get back the integrand. This is a perfect self-check that requires no answer key. If d/dx[F(x) + C] = f(x), you have the right answer. This habit catches the most common errors — wrong constants, sign errors, forgetting the chain rule in reverse.

AM
Dr. Aisha Malik, PhD Mathematics
Senior Lecturer in Applied Mathematics · 12 years teaching calculus
Dr. Malik holds a PhD in Applied Mathematics from the University of Edinburgh and has taught calculus to over 4,000 students. She has reviewed this article for mathematical accuracy and pedagogical clarity.
Technically reviewed by: Prof. James Chen, Stanford Mathematics Department · April 2026