Why do we add + C to every indefinite integral?

Because the derivative of any constant is zero, infinitely many different functions can have the same derivative. For example, both x^3 + 5 and x^3 - 100 have derivative 3x^2. Writing + C captures all of them at once.

Is the indefinite integral the same as the antiderivative?

Essentially, yes. The indefinite integral is the general notation for finding all antiderivatives of a function; it is what you get when you reverse the process of differentiation.

What is the difference between an indefinite and a definite integral?

An indefinite integral (this page) produces a family of functions and includes + C; a definite integral has upper and lower limits and gives a single number. The Fundamental Theorem of Calculus connects the two.

What do I do if I cannot find an antiderivative?

Not every function has an antiderivative in elementary form. Try techniques like u-substitution or integration by parts; if those do not work, use numerical methods or a computer algebra system to approximate or verify the integral.

Indefinite Integral

The indefinite integral finds all functions whose derivative equals a given function, expressed as an antiderivative plus an arbitrary constant.

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

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What each symbol means

What Indefinite Integral takes

Symbol	Meaning
$f$	The integrand—the function you are finding the antiderivative for; if misread as the antiderivative itself, you would confuse what you're starting with versus what you're looking for.
$F$	The antiderivative—the function whose derivative is $f(x)$ ; confusing it with $f$ reverses the entire problem.
$x$	The variable of integration, indicating what quantity the rate of change is with respect to; changing which letter you use as the variable changes what the antiderivative represents.
$C$	The arbitrary constant of integration, accounting for the fact that derivatives of constants are zero; omitting it means missing infinitely many valid antiderivatives.

When to use it

Use the indefinite integral when you need to find all antiderivatives of a function.

Level

Usually taught in: Calculus I

Worked examples

1. Integrate a simple power

Problem

Find

\int 3x^2 dx

$\int 3x^2 dx = 3 \int x^2 dx$
Factor out the constant 3 using the constant multiple rule for integrals.
$= 3 \cdot \frac{x^{3}}{3} + C$
Apply the power rule: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ , with $n = 2$ .
$= x^3 + C$
Simplify by multiplying $3 \cdot \frac{x^3}{3} = x^3$ .

Answer: $x^3 + C$

This uses the power rule directly because $x^2$ is a single power term. We factor out the constant 3, apply the rule to get $x^3/3$ , multiply back by 3 to get $x^3$ , then add the constant of integration.

2. Integrate with fractions and mixed terms

Problem

Find

\int \left(5x^2 - \frac{2}{x^3} + 4\right) dx

$\int \left(5x^2 - 2x^{-3} + 4\right) dx$
Rewrite the fraction $\frac{2}{x^3}$ as $2x^{-3}$ so that the power rule applies to all terms.
$= 5 \int x^2 dx - 2 \int x^{-3} dx + 4 \int dx$
Split the integral into three terms using the linearity (sum rule) of integration.
$= 5 \cdot \frac{x^3}{3} - 2 \cdot \frac{x^{-2}}{-2} + 4x + C$
Apply the power rule to each term: for $x^2$ we get $\frac{x^3}{3}$ , for $x^{-3}$ we get $\frac{x^{-2}}{-2}$ , and for $1$ (the constant) we get $x$ .
$= \frac{5x^3}{3} + x^{-2} + 4x + C$
Simplify: $5 \cdot \frac{x^3}{3} = \frac{5x^3}{3}$ and $-2 \cdot \frac{x^{-2}}{-2} = x^{-2}$ (the negatives cancel).

Answer: $\frac{5x^3}{3} + x^{-2} + 4x + C$

This problem combines polynomial, fractional, and constant terms. Rewriting fractions as negative exponents lets us apply the power rule uniformly. The middle term demonstrates care with signs: when you divide $x^{-3}$ by $-2$ (the new exponent plus 1), the negative signs work out to give a positive $x^{-2}$ term.

3. Find total items prepared at a school bake sale

Problem

At a school bake sale, cupcakes are frosted at a rate of

r(t) = 15t + 12

cupcakes per hour, where

t

is time in hours. If no cupcakes were frosted at

t = 0

, how many cupcakes have been frosted after

t

hours?

$N(t) = \int (15t + 12) dt$
To find the total number of cupcakes frosted (accumulated quantity) from the rate, integrate the rate function.
$= 15 \cdot \frac{t^2}{2} + 12t + C$
Apply the power rule to $15t$ (which is $15t^1$ ) to get $15 \cdot \frac{t^2}{2}$ , and to the constant $12$ to get $12t$ .
$= \frac{15t^2}{2} + 12t + C$
Simplify the first term: $15 \cdot \frac{t^2}{2} = \frac{15t^2}{2}$ .
$N(0) = \frac{15 \cdot 0^2}{2} + 12 \cdot 0 + C = C, \text{ so } C = 0$
Apply the initial condition that $N(0) = 0$ (no cupcakes frosted at time zero) to solve for $C$ .

Answer: $N(t) = \frac{15t^2}{2} + 12t$

The indefinite integral accumulates a rate to find total quantity, which is perfect for a real-world scenario like a bake sale. The arbitrary constant $C$ represents how many cupcakes were already done before counting began. Here it is zero. If $20$ cupcakes had already been frosted, the answer would be $\frac{15t^2}{2} + 12t + 20$ , illustrating why we must include $+ C$ unless an initial condition pins it down.

Common mistakes

Where Indefinite Integral usually goes wrong

The mistake	Why it is wrong	The fix
$\int 3x^2 dx = x^3$	Leaving off the constant $+ C$ gives an incomplete answer, because any two antiderivatives differ only by a constant.	Write $\int 3x^2 dx = x^3 + C$ to represent the entire family of antiderivatives.
$\int x^3 dx = x^4 + C$	The power rule requires you to divide by the new exponent $n + 1 = 4$ ; forgetting this division loses a factor of $\frac{1}{4}$ .	$\int x^3 dx = \frac{x^4}{4} + C$
$\int 5x^2 dx = 5x^3 + C$	After raising $x^2$ to the power $x^3$ , you must divide by $3$ ; the coefficient $5$ gets divided too, giving $\frac{5x^3}{3}$ , not $5x^3$ .	$\int 5x^2 dx = \frac{5x^3}{3} + C$

Tips and when to use something else

Always include $+ C$ at the end: it represents the entire family of antiderivatives, not a single function.
The power rule fails for $\int \frac{1}{x} dx$ because dividing by $n+1 = 0$ is undefined; use the logarithm rule $\int \frac{1}{x} dx = \ln|x| + C$ instead.
Verify your answer by differentiating: if $\frac{d}{dx}[F(x) + C]$ equals the original integrand $f(x)$ , you are correct.
For integrals of $e^x$ , sine, cosine, or other special functions, use their dedicated rules rather than trying to force the power rule.

Related formulas

Power Rule for IntegrationPower Rule for Integration instantly finds antiderivatives of polynomial terms; use it to integrate any power of x except x to the negative one.Fundamental Theorem of CalculusCompute the exact value of a definite integral using the antiderivative—connect area under a curve to the accumulation of change.U-SubstitutionU-substitution simplifies integrals by transforming them into easier forms using a change of variables, ideal when you spot a function and its derivative.Integration by PartsIntegration by Parts is a calculus technique for evaluating integrals of products by transforming them into simpler forms when substitution fails.Definite Integral PropertiesTwo key properties of definite integrals: reversing the bounds negates the integral, and integrating over a single point gives zero.

Frequently asked questions

Why do we add $+ C$ to every indefinite integral?: Because the derivative of any constant is zero, infinitely many different functions can have the same derivative. For example, both $x^3 + 5$ and $x^3 - 100$ have derivative $3x^2$ . Writing $+ C$ captures all of them at once.
Is the indefinite integral the same as the antiderivative?: Essentially, yes. The indefinite integral is the general notation for finding all antiderivatives of a function; it is what you get when you reverse the process of differentiation.
What is the difference between an indefinite and a definite integral?: An indefinite integral (this page) produces a family of functions and includes $+ C$ ; a definite integral has upper and lower limits and gives a single number. The Fundamental Theorem of Calculus connects the two.
What do I do if I cannot find an antiderivative?: Not every function has an antiderivative in elementary form. Try techniques like $u$ -substitution or integration by parts; if those do not work, use numerical methods or a computer algebra system to approximate or verify the integral.

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Reviewed 2026-09-18