Math · Calculus I · Concept

Antiderivatives and indefinite integrals

An antiderivative of f is a function F whose derivative is f. On an interval, every antiderivative of f has the form F(x) + C, written ∫f(x) dx = F(x) + C. Find one by running the power rule and the other derivative rules backward, then check it by differentiating.

Undoing a derivative

Differentiation takes F to f; antidifferentiation goes back from f to F. The answer is never unique: x³, x³ + 5 and x³ − 2 all have derivative 3x². On an interval, any two antiderivatives differ by a constant, so the indefinite integral names the whole family with + C.

\int f (x) d x = F (x) + C means F^{'} (x) = f (x)

\begin{array}{c} \int f (x) d x = F (x) + C \\ means F^{'} (x) = f (x) \end{array}

Run the power rule backward

The derivative of xⁿ⁺¹ is (n + 1)xⁿ. So to antidifferentiate xⁿ, raise the power by one and divide by the new power. The rule fails only for n = −1, where it would divide by zero; that case gives ln|x|.

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C, n \neq - 1

\begin{array}{c} \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C \\ n \neq - 1 \end{array}

Basic antiderivatives

Each entry is a derivative rule read backward. Angles are in radians.

Basic antiderivatives
Function	Antiderivative
k, a constant	kx + C
xⁿ, n ≠ −1	xⁿ⁺¹/(n + 1) + C
1/x	ln\|x\| + C
eˣ	eˣ + C
sin x	−cos x + C
cos x	sin x + C
sec² x	tan x + C

Sums and constant multiples

Antiderivatives work term by term, and a constant factor stays in front, because derivatives behave the same way. There is no such rule for products: ∫x·eˣ dx is not (x²/2)·eˣ + C. A product has to be expanded first, or handled by substitution or integration by parts.

\int (a f + b g) d x = a \int f d x + b \int g d x

\begin{array}{c} \int (a f + b g) d x \\ = a \int f d x + b \int g d x \end{array}

Rewrite roots and reciprocals first

The power rule needs the form xⁿ. Write √x as x^(1/2) and 3/x² as 3x^(−2), then integrate.

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

\begin{array}{c} \int \frac{3}{x^{2}} d x = \int 3 x^{- 2} d x \\ = - 3 x^{- 1} + C \end{array}

Check by differentiating

Differentiate your answer. If you get the integrand back, the antiderivative is right; if not, the difference usually points to a missing divisor or a sign slip. The calculus checker verifies an antiderivative the same way.

An initial condition fixes C

A condition such as f(1) = 5 picks one function out of the family. Motion problems work this way: velocity is an antiderivative of acceleration, position is an antiderivative of velocity, and the starting values fix the constants.

Common mistakes

Leaving off + C: an indefinite integral is a family of functions, not a single function.
Using the power rule on 1/x, which would divide by zero. The antiderivative of 1/x is ln|x| + C.
Getting the sign wrong: the antiderivative of sin x is −cos x + C, because the derivative of cos x is −sin x.
Integrating a product factor by factor: ∫x·eˣ dx is not (x²/2)·eˣ + C. Differentiating that guess with the product rule does not give x·eˣ.
Forgetting to divide by the new exponent: ∫x^(1/2) dx is (2/3)x^(3/2) + C, not x^(3/2) + C.

Key terms

Antiderivative: A function whose derivative is the given function, such as x³ for 3x². Any two antiderivatives on one interval differ by a constant, which is why answers carry + C.
Indefinite integral: ∫f(x) dx, the whole family of antiderivatives of f, written with + C, such as ∫2x dx = x² + C. It has no limits of integration.
Constant of integration: The + C added to an antiderivative, because the derivative of any constant is 0. On a domain split into separate pieces, each piece can have its own constant.
Integrand: The function being integrated: the expression whose values contribute to the accumulation. In ∫f(x) dx, f(x) is the integrand and x is the integration variable.
Initial-value problem: A differential equation together with a condition that picks out one solution, such as f′(x) = 3x² + 2 with f(1) = 5. For an antiderivative, the condition determines the constant C.

Work through an example

Find ∫(6x² − 4x + 5) dx and check the answer by differentiating.

Find an indefinite integral term by term →

Solve an initial-value problem →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

OpenStax Calculus Volume 1 — Antiderivatives

Make it concrete

Try in the workspace

Open the example inputs, change a value and keep a useful result on your board.

Verify it in the antiderivative checker Check the antiderivative in Math Open worked example on a board Antiderivative power rule in Math Reference

Your existing work stays on this device. Examples open as editable copies.