Antiderivative Calculator: Power Rule and Polynomial Guide (2026)

Antiderivative calculator: power rule, polynomials, exponential, and log rules with step-by-step worked examples. Free, no signup.

You have 6x² in front of you and need to reverse the derivative. The power rule for antiderivatives handles this: add one to the exponent, then divide by that new exponent. For 6x², that gives 6x³/3, which simplifies to 2x³. The Antiderivative Calculator runs this for any function instantly, but understanding the pattern lets you handle composite functions and verify results without a tool.

The antiderivative (also called the indefinite integral) is the reverse of differentiation. If f'(x) = 6x², then f(x) = 2x³ + C. The constant C is not optional: it represents a family of functions that all share the same derivative. Skipping it is a common error that changes the mathematical meaning of the answer.

This guide covers the power rule, polynomial antiderivatives, exponential and logarithm cases, u-substitution for composite functions, and the most frequent mistakes. Trig function antiderivatives have their own set of identities and are covered separately.

The Power Rule for Antiderivatives

The power rule is the most-used formula in basic integration:

∫ xⁿ dx = x^(n+1) / (n+1) + C     where n ≠ -1

Add one to the exponent, divide the coefficient by that new exponent. Two operations, applied in that order.

Worked examples:

• ∫ x⁴ dx: exponent becomes 5, divide 1 by 5: x⁵/5 + C
• ∫ 3x² dx: exponent becomes 3, divide 3 by 3: x³ + C
• ∫ x⁻³ dx: exponent becomes −2, divide 1 by −2: −x⁻²/2 + C
• ∫ √x dx = ∫ x^(1/2) dx: exponent becomes 3/2, divide by 3/2: (2/3)x^(3/2) + C

The rule works for negative exponents and fractional exponents as long as n ≠ −1. The n = −1 case is ∫ 1/x dx = ln|x| + C, a separate rule covered below.

Constants: ∫ k dx = kx + C. A constant k is effectively k · x⁰, and applying the power rule gives k · x¹/1 = kx.

Polynomials: Applying the Power Rule Term by Term

Polynomials are antidifferentiated term by term. Apply the power rule to each term separately, then combine.

Example: ∫ (4x³ − 6x² + 2x − 5) dx

• ∫ 4x³ dx = 4 · x⁴/4 = x⁴
• ∫ −6x² dx = −6 · x³/3 = −2x³
• ∫ 2x dx = 2 · x²/2 = x²
• ∫ −5 dx = −5x

Result: x⁴ − 2x³ + x² − 5x + C

Only one C is needed regardless of the number of terms. Each term technically produces its own constant, but they all collapse into a single C.

Verification by differentiation: Take the derivative of x⁴ − 2x³ + x² − 5x. You get 4x³ − 6x² + 2x − 5, which matches. This is the fastest check for any antiderivative: differentiate it and confirm you get the original function back.

Fractional coefficients are treated the same way. For ∫ (x²/3 + x/4) dx:

• ∫ x²/3 dx = (1/3) · x³/3 = x³/9
• ∫ x/4 dx = (1/4) · x²/2 = x²/8

Result: x³/9 + x²/8 + C

Antiderivative Rules for Exponential and Log Functions

Three rules handle the most common non-polynomial cases:

∫ eˣ dx = eˣ + C
∫ aˣ dx = aˣ / ln(a) + C     (a > 0, a ≠ 1)
∫ 1/x dx = ln|x| + C

The function eˣ is its own antiderivative. That is not a coincidence: e is defined so that this property holds, which is why it appears throughout calculus, differential equations, and complex analysis.

For base-a exponentials:

∫ 2ˣ dx = 2ˣ / ln(2) + C. Since ln(2) ≈ 0.693, this is approximately 2ˣ / 0.693 + C. For base 10: ∫ 10ˣ dx = 10ˣ / ln(10) + C ≈ 10ˣ / 2.303 + C.

For the log rule:

∫ 1/x dx = ln|x| + C. The absolute value is required when x can be negative, because ln(x) is only defined for positive x. If the problem specifies a domain of positive x, ln(x) + C is acceptable. Exams that require full generality need the absolute value.

This is the case the power rule cannot handle. Applying ∫ x⁻¹ dx with the power rule gives x⁰/0, which is undefined. The ln|x| result stands on its own.

The same relationship between differentiation and antidifferentiation extends to multivariable functions, which the Partial Derivative Calculator handles when you need to differentiate in one variable while treating others as constants.

U-Substitution: When the Power Rule Does Not Apply Directly

U-substitution handles integrals of composite functions. The method reverses the chain rule from differentiation.

When to use it: Look for a function and its derivative both present in the integrand. The pattern: ∫ f(g(x)) · g'(x) dx.

Example: ∫ 2x(x² + 1)⁵ dx

Let u = x² + 1. Then du/dx = 2x, so du = 2x dx.

∫ 2x(x² + 1)⁵ dx = ∫ u⁵ du = u⁶/6 + C = (x² + 1)⁶/6 + C

The 2x and dx pair becomes du exactly, so the substitution works cleanly.

Example: ∫ cos(3x) dx

Let u = 3x. Then du = 3 dx, so dx = du/3.

∫ cos(3x) dx = ∫ cos(u) · du/3 = (1/3) sin(u) + C = (1/3) sin(3x) + C

When it fails: If the derivative of your chosen u is absent from the integrand, u-substitution will not simplify the integral. ∫ (x² + 1)⁵ dx does not yield (x² + 1)⁶/6 + C — the 2x needed for du is missing. Expanding the binomial or applying a different technique is required.

The Constant of Integration: Why It Is Never Optional

Every indefinite integral includes + C. This is not a formality and cannot be dropped.

The constant represents an infinite family of antiderivatives. The functions x³ + 5, x³ − 7, and x³ are all valid antiderivatives of 3x². They differ only by a constant, and all share the same derivative. Without C, you are implying there is one antiderivative, which is incorrect.

Solving for C from initial conditions:

When a specific point on f(x) is known, you can find the particular antiderivative.

Given f'(x) = 3x² and f(0) = 5:

f(x) = x³ + C
f(0) = 5: 0³ + C = 5 → C = 5
Particular solution: f(x) = x³ + 5

This function passes through (0, 5). The general form x³ + C covers every vertical shift of x³, while x³ + 5 is the one that satisfies the initial condition.

In definite integrals, C cancels out during evaluation between the bounds. For indefinite integrals, which is what "find the antiderivative" always means, C must appear in the final answer.

For the complete set of antiderivative rules for sin, cos, tan, sec, csc, and cot, the Antiderivative of Trig Functions guide covers each identity with worked examples.

Checking Your Antiderivative: Differentiation as Verification

Every antiderivative can be verified in under a minute. Differentiate your answer. If the result matches the original integrand, the antiderivative is correct.

For ∫ 3x² dx = x³ + C: differentiate x³ + C to get 3x². Matches. Correct.

For the polynomial from the section above: differentiate x⁴ − 2x³ + x² − 5x + C. The result is 4x³ − 6x² + 2x − 5, which is the original integrand. Correct.

This check exposes the errors that are hardest to catch by inspection:

Coefficient errors after the power rule. If you wrote ∫ 5x⁴ dx = 5x⁵ + C instead of x⁵ + C, differentiating 5x⁵ gives 25x⁴, not 5x⁴. The mismatch is immediate.

Absolute value dropped in ln|x|. The derivative of ln|x| is 1/x for all x ≠ 0. If you wrote ln(x) and the domain includes negative values, the function is undefined there. The verification step forces you to confirm the domain is handled correctly.

Back-substitution errors after u-substitution. Once you substitute back from u to x, differentiate the full result. Any constant placed in the wrong position or incorrect chain rule reversal surfaces when the derivative does not match the original integrand.

Verification is not optional for exam problems. It is the same cost as re-checking your arithmetic, and it is more reliable because it tests the structure of the answer, not just the arithmetic.

The Math Calculators section covers antiderivatives, derivatives, matrix operations, and statistical tools in one place.