Indefinite Integrals

Antiderivatives

The Constant of Integration

Indefinite Integral Notation

Basic Antiderivative Formulas

Linearity of Indefinite Integrals

Verifying Antiderivatives

Connection to Definite Integrals

Summary: The Indefinite-Integration Workflow

Common Pitfalls

Indefinite Integrals FAQ

Reversing Differentiation

Differentiation takes a function and produces its rate of change. The indefinite integral reverses this process: given a rate of change, find the original function.

If

F'(x) = f(x)

, then

F

is an antiderivative of

f

. The indefinite integral collects all such antiderivatives:

Summary: The Indefinite-Integration Workflow

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

The constant

C

reflects a fundamental ambiguity. Since the derivative of any constant is zero, functions differing by a constant share the same derivative. The family

F(x) + C

captures every function whose derivative equals

f(x)

.

Finding antiderivatives is the central challenge. Unlike differentiation, which follows mechanical rules, integration demands pattern recognition, technique, and sometimes ingenuity.

Key Terms

Antiderivative— a function whose derivative is

f

Indefinite Integral—

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

, the full family

Integrand— the function

f(x)

being integrated

Definite Integral— connected by the Fundamental Theorem

See All Calculus Definitions →

Antiderivatives

A function

F

is an antiderivative of

f

if:

Antiderivatives

F'(x) = f(x)

The antiderivative reverses differentiation. Given

f(x) = 2x

, the function

F(x) = x^2

is an antiderivative because

(x^2)' = 2x

.

But

F(x) = x^2 + 5

is also an antiderivative—its derivative is likewise

2x

. So is

x^2 - 17

, or

x^2 + \pi

. Any function of the form

x^2 + C

differentiates to

2x

RULE

Antiderivatives are not unique. They form a family of functions, all differing by constants.

F(x) = ∫₀ˆx t² dt = x³/3 is an antiderivative of x²

The lower curve is f(t) = t² and the upper one is the area accumulated under it from 0; the slope of the upper curve at any x equals the height of the lower one there, which is what F′ = f means. Start the accumulation from a different a and the upper curve shifts by a constant: the + C of the indefinite integral. Change the starting point and watch the whole antiderivative slide on the fundamental theorem of calculus visualizer.

Every antiderivative of f is one of these accumulation curves, shifted.

The Constant of Integration

F(x)

is one antiderivative of

f(x)

, then every antiderivative has the form:

Antiderivative Family

\int f(x)\, dx = F(x) + C \quad \text{where } F'(x) = f(x)

Learn more about this formula: Antiderivative Family →

Browse all integral formulas →

where

C

is an arbitrary constant. This follows from a basic fact: if two functions have the same derivative on an interval, they differ by a constant.

The "

+ C

" in indefinite integrals is not optional notation—it represents the complete answer. Omitting it gives only one member of the family when infinitely many exist.

Initial conditions pin down

C

. If you know that

F(0) = 3

, for instance, you can solve for the specific constant that satisfies this requirement.

Indefinite Integral Notation

Notation

Indefinite Integral Notation

The three-part symbol, the constant that is not optional, and the case convention that carries the whole subject.

dx

as a differential — differentials; the prime — the derivative function.

\int f(x)\, dx

The indefinite integral of f of x, d x

Leibniz's elongated S, for summa. No limits on the sign — indefinite: the output is a family of functions, not a number.

f(x)

is the integrand;

dx

names the variable and closes it off. The mathematical keyboard carries

\int

CasesWith limits attached it becomes the definite integral — a number, a different object, same glyph. Under substitution the

dx

is a live differential:

du = g'(x)\,dx

is real algebra, not bookkeeping.

Also written

\int dx\, f(x)

— differential first, standard in physics texts, where the variable announces itself before a long integrand.

Do not confuse

dx

as decoration. Dropping it leaves the variable unnamed —

\int 2t\,dx

and

\int 2t\,dt

are different answers.

F(x) + C

F of x, plus C

The general antiderivative. “

+C

” is part of the answer, not a garnish — The Constant of Integration above insists on it. One symbol, ranging over all reals, carries the entire family.

CasesSums take a single

C

— the pieces' constants merge on arrival. For

\int \frac{1}{x}\,dx = \ln|x| + C

the honest version has a separate constant on each side of

0

; the single

C

is the standard abbreviation.

Also written

K

, or

c_1

, when

C

is spoken for — any letter declared arbitrary.

Do not confuseThe critical number

c

of graph analysis — one fixed, found point; this

C

is deliberately free. In definite integrals it cancels, which is why it never appears there.

F'(x) = f(x)

F prime equals f

The case convention: a capital letter names an antiderivative of its own lowercase. The letter case is the claim — writing

F

asserts

F' = f

with no further words.

Cases

G/g

H/h

work identically. The pairing carries into the Fundamental Theorem,

\int_a^b f(x)\,dx = F(b) - F(a)

— Connection to Definite Integrals below.

Also written

F = \int f

— operator shorthand, bare sign and no variables, common on blackboards.

Do not confuse“

F

is the antiderivative.”

F

is one member — the family is

F + C

, and uniqueness fails by exactly a constant.

All calculus symbols →↑ Integrals

Basic Antiderivative Formulas

Several antiderivatives appear constantly and should be memorized.

Power rule (for

n \neq -1

Basic Antiderivative Formulas

\int x^n\, dx = \frac{x^{n+1}}{n+1} + C

Reciprocal:

Reciprocal Antiderivative

\int \frac{1}{x}\, dx = \ln|x| + C

Learn more about this formula: Reciprocal Antiderivative →

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Pitfall: Using the power rule on 1/x

Writing

\frac{1}{x}

x^{-1}

and applying the power rule gives

\frac{x^{0}}{0}

, a division by zero. The power rule excludes

n = -1

; the antiderivative is

\ln|x| + C

The reciprocal formula fills the one gap in the power rule. Its antiderivative is a logarithm, not a power of

x

, and the absolute value lets one formula serve both sides of

0

: for

x > 0

(\ln x)' = \frac{1}{x}

, and for

x < 0

(\ln(-x))' = \frac{-1}{-x} = \frac{1}{x}

. At

x = 0

itself neither

\frac{1}{x}

nor

\ln|x|

is defined.

Exponential:

Exponential Antiderivative

\int e^x\, dx = e^x + C

Learn more about this formula: Exponential Antiderivative →

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Trigonometric:

Basic Antiderivative Formulas

\int \cos x\, dx = \sin x + C \qquad \int \sin x\, dx = -\cos x + C

Pitfall: Losing the minus sign in the sine integral

Writing

\int \sin x\, dx = \cos x + C

gives a function whose derivative is

-\sin x

, not

\sin x

. The correct antiderivative is

-\cos x + C

Both trigonometric formulas are derivatives read backwards, which is also the fastest way to recall them:

(\sin x)' = \cos x

and

(-\cos x)' = \sin x

. A quick derivative of a candidate answer settles which function goes with which, and with which sign, without memorizing a separate table.

The special integrals page provides a more complete list.

Integrand f(x)	∫ f(x) dx
xⁿ (n ≠ −1)	xⁿ⁺¹ / (n + 1) + C
1 / x	ln \|x\| + C
e^x	e^x + C
cos x	sin x + C
sin x	−cos x + C

Linearity of Indefinite Integrals

Indefinite integrals obey the same linearity rules as definite integrals.

Sum rule:

Linearity of Indefinite Integrals

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

Constant multiple rule:

Constant Multiple Rule (Integrals)

\int c \cdot f(x)\, dx = c \int f(x)\, dx

Learn more about this formula: Constant Multiple Rule (Integrals) →

Browse all integral formulas →

These rules reduce complex integrands to combinations of simpler ones. For example:

Linearity of Indefinite Integrals

\int (3x^2 + 5x - 2)\, dx = 3 \cdot \frac{x^3}{3} + 5 \cdot \frac{x^2}{2} - 2x + C = x^3 + \frac{5x^2}{2} - 2x + C

Pitfall: Pulling a variable out of the integral

Moving

x

out of

\int x \cdot 2x\, dx

as if it were a constant gives

x \cdot x^2 + C = x^3 + C

, whose derivative is

3x^2

, not

2x^2

. Multiplying first,

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

In the example, the coefficients

3

5

and

-2

are fixed numbers, so each one passes through the integral sign unchanged, and what is left in each term is a power of

x

that the power rule handles. The same two steps, splitting the sum and then taking out the numbers, turn any polynomial into a list of power-rule integrals with a single

+ C

at the end.

Verifying Antiderivatives

Integration has a built-in check: differentiate your answer.

If

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

, then

F'(x)

must equal

f(x)

. If it doesn't, an error occurred.

For example, suppose you compute:

Antiderivative of Secant Squared

\int \sec^2 x\, dx = \tan x + C

Learn more about this formula: Antiderivative of Secant Squared →

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Verify:

(\tan x)' = \sec^2 x

. Correct.

This check catches sign errors, missing constants, and algebraic mistakes. It works because differentiation is mechanical—once you have a candidate antiderivative, verification is straightforward.

Connection to Definite Integrals

Indefinite and definite integrals serve different purposes but are linked by the Fundamental Theorem of Calculus.

The indefinite integral finds antiderivatives:

Connection to Definite Integrals

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

The definite integral uses an antiderivative to compute accumulated quantity:

Connection to Definite Integrals

\int_a^b f(x)\, dx = F(b) - F(a)

COROLLARY

The constant

C

cancels when evaluating

F(b) - F(a)

, so any antiderivative works.

This connection, detailed on the rules page, is why mastering indefinite integration enables computation of definite integrals.

Why the + C drops out of a definite integral

x² and x² + 2 are two antiderivatives of 2x, one sitting 2 units above the other. Between a = 1 and b = 2 the lower curve rises from 1 to 4 and the upper from 3 to 6: the same rise of 3 (the two columns). Shifting a curve up moves both endpoints by the same amount, so F(b) − F(a) is the same for every member of the family and equals ∫₁² 2x dx = 3.

That is why a definite integral needs no + C, while an indefinite one does.

Summary: The Indefinite-Integration Workflow

Every indefinite integral on this page follows the same five-step procedure, whether the integrand is a single power function or a complicated composition that requires a technique. The table below collects the workflow with a pointer to the sibling page where each step's detail lives — special integrals for the formula library, rules for linearity and the FTC, techniques for transformations. Use this as a reminder while working a problem: every step matters, and skipping the final verification is the single most common source of errors.

Step	Action	Where the detail lives
1	Recognize the form — does the integrand match a known antiderivative formula?	special integrals
2	Apply linearity — split sums and factor out constants to reduce to known pieces	integration rules
3	Compute the antiderivative F(x) — apply a formula directly, or use a technique to transform the integrand first	techniques
4	Add the constant of integration — the answer is F(x) + C, never just F(x)	section 2 above (constant of integration)
5	Verify by differentiating — F'(x) should return the original integrand f(x)	section 6 above (verifying antiderivatives)

Common Pitfalls

Four mistakes cause most wrong answers in basic indefinite integration. Three of them are marked where they happen in the sections above; the first is the one the section on the constant of integration warns against. Here each is worked through in full, with the wrong and the right working side by side.

Forgetting the + C

The antiderivative is found correctly, but the answer is written as

F(x)

alone. That names one antiderivative where the indefinite integral asks for all of them, and it leaves nothing to adjust when an initial condition has to be met.

Wrong working:

\int 2x\, dx = x^2

. With the condition

F(0) = 3

, this gives

F(0) = 0^2 = 0

, and there is no constant left to fix it.

Right working:

\int 2x\, dx = x^2 + C

. The condition gives

0^2 + C = 3

, so

C = 3

and

F(x) = x^2 + 3

.

Why it happens: differentiation wipes out constants, so the constant is invisible when the derivative is read backwards. In definite integrals the constant cancels and is never written, and the habit carries over.

How to check: every indefinite integral has exactly one arbitrary constant. Before an initial condition is used, there must be a

C

in the answer to solve for.

Using the power rule on 1/x

The integrand

\frac{1}{x}

is rewritten as

x^{-1}

and the power rule is applied as for any other power. With

n = -1

the rule's denominator

n + 1

0

.

Wrong working:

\int x^{-1}\, dx = \frac{x^{-1+1}}{-1+1} + C = \frac{x^{0}}{0} + C

, which is undefined.

Right working: the power rule holds only for

n \neq -1

. The reciprocal formula covers this case:

\int \frac{1}{x}\, dx = \ln|x| + C

.

Why it happens:

x^{-1}

looks like every other power of

x

, and the condition

n \neq -1

stands beside the rule, where it is easy to skip.

How to check: differentiate.

(\ln|x|)' = \frac{1}{x}

for every

x \neq 0

. No power of

x

can work:

(x^k)' = kx^{k-1}

has the exponent

-1

only when

k = 0

, and then the derivative is

0

.

Losing the minus sign in the sine integral

The antiderivatives of

\sin x

and

\cos x

are recalled from the derivative table, and the minus sign ends up on the wrong formula or disappears.

Wrong working:

\int \sin x\, dx = \cos x + C

, or

\int \cos x\, dx = -\sin x + C

.

Right working:

\int \sin x\, dx = -\cos x + C

, because

(-\cos x)' = -(-\sin x) = \sin x

. And

\int \cos x\, dx = \sin x + C

, because

(\sin x)' = \cos x

.

Why it happens: in the derivative table the minus sign belongs to the cosine,

(\cos x)' = -\sin x

. Read backwards, the sign moves to the other formula, and it is easy to leave it where it was.

How to check: differentiate the answer.

(\cos x)' = -\sin x

has the wrong sign. A definite value also catches it:

\int_0^{\pi} \sin x\, dx

is the area under one arch of the sine, which is positive. With

-\cos x

it is

-\cos \pi + \cos 0 = 2

; with

\cos x

it would be

\cos \pi - \cos 0 = -2

.

Pulling a variable out of the integral

The constant multiple rule lets a number pass through the integral sign. A factor that contains

x

is moved out the same way, and only the rest is integrated.

Wrong working:

\int x \cdot 2x\, dx = x \int 2x\, dx = x \cdot x^2 + C = x^3 + C

.

Right working: multiply first,

x \cdot 2x = 2x^2

, then

\int 2x^2\, dx = 2 \cdot \frac{x^3}{3} + C = \frac{2x^3}{3} + C

.

Why it happens: in

\int c \cdot f(x)\, dx

the constant stands in front of the function, just where a variable factor would stand, and the condition that

c

does not change with

x

is easy to overlook.

How to check: differentiate.

(x^3)' = 3x^2

, not

2x^2

;

\left(\frac{2x^3}{3}\right)' = 2x^2

Indefinite Integrals FAQ

Why do all antiderivatives of a function differ by a constant?

Because two functions with the same derivative on an interval can only differ by a constant. If F′ = G′ then (F − G)′ = 0, and a function whose derivative is zero everywhere on an interval is constant. So once you find one antiderivative F, the whole family is F + C — nothing outside that form can have the right derivative.Read more →

Do you need a separate + C for each term?

No — one constant covers the whole answer. Integrating term by term, each piece contributes its own arbitrary constant, but a sum of arbitrary constants is just another arbitrary constant, so they merge into a single C. Writing C₁ + C₂ + C₃ is not wrong, only redundant. Collapse them and write + C once, at the end.Read more →

How do you find the value of C?

You need one known value of the antiderivative — an initial condition. Integrate to get F(x) + C, substitute the known point, and solve. If ∫2x dx = x² + C and you are told F(0) = 3, then 0² + C = 3, so C = 3 and F(x) = x² + 3. Without such a condition, C stays arbitrary.Read more →

Does the dx in an integral actually matter?

Yes — it names the variable of integration, and dropping it leaves the expression ambiguous. ∫2t dx and ∫2t dt have different answers: the first treats t as a constant and gives 2tx + C, the second integrates in t and gives t² + C. The dx is also a live differential under substitution.Read more →

Why is an antiderivative written with a capital F?

It is a case convention that carries a claim: writing F for an antiderivative of f asserts F′ = f without saying so in words. The same pairing works for G and g, or H and h. It is what makes the Fundamental Theorem readable as F(b) − F(a) — the capital signals which function is the antiderivative.Read more →