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Math · Calculus

Integral Calculator

Find the indefinite integral of a polynomial with the reverse power rule.

Polynomials only — e.g. 6x^2 + 2x - 5.
∫ f(x) dx
2x^3 + x^2 − 5x + C

Reverse power rule, plus the constant of integration.

Integration reverses the power rule: ∫a·xⁿ dx = a ÷ (n+1) · xⁿ⁺¹ + C — raise the exponent by one and divide by it. So ∫(6x − 5) dx = 3x² − 5x + C, whose derivative is 6x − 5. The “+ C” is required for indefinite integrals.

The reverse power rule

Integration undoes differentiation. Where the power rule lowers an exponent, the reverse power rule raises it: add one to the exponent, then divide by that new exponent. Because many functions share the same derivative, an indefinite integral always ends with a constant of integration, “+ C”.

∫ a·xⁿ dx = a ÷ (n + 1) · xⁿ⁺¹ + C

raise the exponent by one, divide by the new exponent; valid for n ≠ −1

Worked example

Integrate ∫ (6x − 5) dx:

  1. 1
    Integrate 6x (here n = 1). Raise to x²; divide by the new exponent 2: 6 ÷ 2 · x² = 3x².
  2. 2
    Integrate −5 (here n = 0). Raise to x¹; divide by 1: −5 ÷ 1 · x = −5x.
  3. 3
    Add the constant. Result: 3x² − 5x + C. Differentiating it gives back 6x − 5. ✓

Common integral rules

The standard antiderivatives; every indefinite integral adds + C.

Function f(x)Integral ∫ f(x) dx
xⁿ (n ≠ −1)xⁿ⁺¹ ÷ (n + 1) + C
1 ÷ xln|x| + C
eˣ + C
cos xsin x + C
sin x−cos x + C
constant kk·x + C

The “+ C” and the n = −1 exception

Why “+ C”? Any two antiderivatives of the same function differ only by a constant — the derivative of a constant is zero — so an indefinite integral represents a whole family of curves.

The rule breaks at n = −1. You can’t divide by n + 1 = 0. The integral of 1 ÷ x is instead ln|x| + C, a special case worth memorising.

Definite integrals. To find an area, evaluate the antiderivative at the upper and lower limits and subtract; the constant C cancels out. This tool returns the indefinite integral.

What functions are supported?
Polynomials in x. For trig, exponential, and 1/x integrals, see the rules table to do them by hand.
Why the “+ C”?
An indefinite integral has infinitely many antiderivatives differing by a constant, so C stands in for that unknown constant.
Why doesn’t the rule work for 1/x?
That is the n = −1 case, where dividing by n + 1 = 0 fails. Its integral is ln|x| + C instead.
How can I check an integral?
Differentiate your answer — you should get back the original function. ∫(6x − 5)dx = 3x² − 5x + C, and d/dx of that is 6x − 5.
Definite integrals?
Not yet — this returns the indefinite integral. For a definite one, evaluate the antiderivative at both limits and subtract.
How do I integrate a constant term?
Treat it as k·x⁰, so it becomes k·x + C. For example ∫5 dx = 5x + C, since the derivative of 5x is 5.