Every differentiation rule read backwards is an integration rule, and the indefinite integral is where that reversal happens: given a rate of change, it recovers the family of functions that could have produced it. It is the essential first half of almost all integral calculus, and the step where most students get stuck — not because any single rule is hard, but because choosing which rule applies takes practice. This calculator finds antiderivatives directly, so you can confirm your technique, unblock a problem, and study the shape of a correct answer.
Quick Answer
The indefinite integral is written ∫ f(x) dx = F(x) + C, where F is an antiderivative of f and C is an arbitrary constant. The answer is always a family of functions, never a single one, because the derivative of any constant is zero.
Why the + C Is Not Decoration
Differentiate x³ and you get 3x². Differentiate x³ + 7, or x³ − 40, and you still get 3x². Integration cannot tell which constant was there originally, so it restores all possibilities at once with + C. Graphically, the antiderivative of a function is a stack of identical curves shifted vertically; C picks which one. In applications, an initial condition — a starting value you know — is what pins C down to a specific number.
The basic power rule runs the familiar differentiation rule in reverse: ∫ xⁿ dx = xⁿ⁺¹ / (n + 1) + C, valid for every n except −1, whose integral is ln|x| instead. Around that core sit the standard forms worth recognising on sight: ∫ eˣ dx = eˣ + C, ∫ sin x dx = −cos x + C, ∫ cos x dx = sin x + C, ∫ sec²x dx = tan x + C, and ∫ 1/x dx = ln|x| + C. Most textbook integrals are these forms dressed up with coefficients and inner functions, which is why the real skill is pattern recognition: strip away the decoration, name the basic form, then handle the decoration with substitution or parts.
How to Use the Indefinite Integral Calculator
- Enter the integrand exactly as written, using brackets to make grouping unambiguous.
- Check that the variable in your expression matches the variable you are integrating with respect to.
- Run the calculation and study the antiderivative's structure before worrying about the + C — identify which basic form or technique the answer came from.
- Verify the result by differentiating it mentally (or with a derivative tool) and confirming you get your original function back.
- If the answer uses a form you did not expect, simplify your own attempt — many correct antiderivatives look different yet differ only by a constant.
- When a problem supplies an initial condition, substitute it into your general answer to solve for C before declaring the result final.
A Worked Example: ∫ (6x² − 4x + sec²x) dx
Integrate term by term. The integral of 6x² is 6 · x³/3 = 2x³. The integral of −4x is −4 · x²/2 = −2x². The integral of sec²x is tan x, straight from the standard table. Assembling the pieces: 2x³ − 2x² + tan x + C. Verify by differentiating: 6x² − 4x + sec²x — the original integrand returns exactly, which is the definitive test of any antiderivative. Notice there is only one + C for the whole answer, not one per term: the constants from each term merge into a single arbitrary constant. That detail is small, but examiners look for it.
Antidifferentiation Mistakes That Recur Every Term
Forgetting + C is the classic, and it genuinely matters once initial conditions appear. The second is applying the power rule to 1/x and writing x⁰/0 nonsense instead of ln|x| — and dropping the absolute value around the logarithm. A third is the reverse-chain-rule slip: ∫ (2x + 1)⁵ dx is not (2x + 1)⁶/6, because the inner derivative 2 must be divided out, giving (2x + 1)⁶/12. When an answer looks different from the mark scheme, do not panic — differentiate both; if both return the integrand, both are correct and differ by a constant. For integrals that resist the basic forms, the U Substitution Calculator handles the change-of-variable technique, and once your antiderivative is solid, the Definite Integral Calculator evaluates it between limits.
Pinning Down C with an Initial Condition
The + C stops being abstract the moment a problem tells you one point the antiderivative must pass through. Suppose a particle's velocity is v(t) = 3t² − 4t and its position at t = 0 is s(0) = 5. Integrating term by term, the position is s(t) = t³ − 2t² + C. Now impose the condition: s(0) = 0 − 0 + C = 5, so C = 5 and the specific position function is s(t) = t³ − 2t² + 5. At t = 2, the particle sits at 8 − 8 + 5 = 5 — back at its start, which the velocity's sign change (v is negative until t = 4/3) makes plausible: it drifted backwards, then returned. This two-step pattern — general antiderivative first, initial condition second — is the standard opening of every differential-equations problem, and it is the practical reason the constant is carried through every line of working instead of being tacked on at the end. Dropping it mid-calculation and re-adding it later works only if nothing in between depends on its value, an assumption that fails exactly when the problem is interesting.
A final micro-example shows how the standard forms combine under a single integral sign: ∫ (1/x + 1/√x) dx. The first term is the exceptional one, ln|x|. The second is a power in disguise: 1/√x = x^(−1/2), whose integral is x^(1/2)/(1/2) = 2√x. Together: ln|x| + 2√x + C, verified instantly by differentiating — 1/x + 1/√x returns. Rewriting roots and reciprocals as powers before integrating, and treating 1/x as the lone exception to the power rule, resolves a large share of the integrals students initially judge to be hard.
The Antiderivative Is a Means, Not an End
Nearly everything in integral calculus passes through this step: areas, volumes, differential equations, probability densities. Getting fluent here — and checking your fluency with this calculator rather than trusting a first attempt — is the highest-leverage habit in the subject. For a broader set of integration techniques, the general Integral Calculator covers the methods beyond the basic table.