Integral Calculator

Integral calculator that shows the reverse power rule working

Integrate polynomial expressions of any degree, term by term, with the exponent being raised and the division being performed in front of you rather than collapsed into a result. The constant of integration is included, because leaving it out is what loses marks.

Part of SolveStep's calculus solver. To check any antiderivative, differentiate it in the derivative calculator.

Polynomial integration by the reverse power rule, with the constant of integration included

The reverse power rule

Integration undoes differentiation, and the + C is not optional

Integrating is differentiating in reverse. The power rule for derivatives brings the exponent down and reduces it; the power rule for integrals does the opposite, raising the exponent by one and dividing by the new exponent. ∫x² dx = x³/3 + C, and differentiating x³/3 gives x² back, which is the check worth running on every answer.

The + C is the part students drop and markers deduct for. Differentiating destroys constants: x³/3, x³/3 + 7 and x³/3 − 1000 all have the same derivative. Running the process backwards therefore cannot recover which constant was there, so every indefinite integral stands for a whole family of functions, and the C is what says so.

There is one exponent the rule cannot handle. Raising −1 by one gives 0, and dividing by zero is impossible, so ∫x⁻¹ dx is not x⁰/0. It is ln|x| + C, which is a separate result you memorise rather than derive from the power rule.

Reverse power rule

∫xⁿ dx = xⁿ⁺¹/(n+1) + C

Add one to the exponent, then divide by the new exponent. Valid for every n except −1.

Why + C

d/dx[x³/3] = d/dx[x³/3 + 7]

Both give x². The constant vanished when differentiating, so integration cannot recover it.

The exception

∫(1/x) dx = ln|x| + C

The power rule would need division by zero here, so this case is its own separate result.

Definite integrals

∫₀ⁱ f = F(b) − F(a)

Antidifferentiate, evaluate at both limits, subtract. The C cancels, which is why definite integrals never carry one.

Method

How to integrate a polynomial, step by step

Split the expression into terms.

Integration distributes over addition and subtraction just as differentiation does, so ∫(3x² + 2x − 5) dx is three separate integrals added together.

Pull any constant multiplier outside.

∫3x² dx = 3∫x² dx. The constant waits outside while the power rule runs, then multiplies back in at the end.

Add one to the exponent.

x² becomes x³. A bare x is x¹, so it becomes x². A constant like 5 is 5x⁰, so it becomes 5x¹, which is why constants integrate to a multiple of x.

Divide by the new exponent.

x³ becomes x³/3. Combined with the constant that was waiting outside, ∫3x² dx = 3 · x³/3 = x³.

Add the constant of integration, then check by differentiating.

Write + C, then differentiate your answer. If it does not return the original expression exactly, the arithmetic went wrong somewhere and you have just caught it for free.

Worked examples

Indefinite, definite, and the exception

A polynomial, term by term

∫(3x² + 2x − 5) dx

3x²: raise to x³, divide by 3, giving 3x³/3 = x³.
2x: raise to x², divide by 2, giving 2x²/2 = x².
−5: this is −5x⁰, so it becomes −5x.
Add + C, then check: d/dx(x³ + x² − 5x) = 3x² + 2x − 5, which is the original.

x³ + x² − 5x + C

A definite integral

∫ from 0 to 2 of x² dx

Antidifferentiate first, ignoring the limits: x³/3.
Evaluate at the upper limit: 2³/3 = 8/3.
Evaluate at the lower limit: 0³/3 = 0.
Subtract: 8/3 − 0 = 8/3. No + C appears, because it would be added at both limits and cancel in the subtraction.

8/3

The exponent the rule cannot take

∫(1/x) dx

Written as a power this is ∫x⁻¹ dx.
The rule says add one to the exponent, which gives x⁰, then divide by the new exponent, which is 0.
Dividing by zero is impossible, so the power rule simply does not apply to this one case.
The correct result is ln|x| + C. The absolute value matters: 1/x is defined for negative x too, and ln alone is not.

ln|x| + C

Integrate your own expression above →

Scope

What this calculator integrates, and what it does not

Integration here covers polynomial expressions of any degree, term by term, with the constant of integration included. That covers the overwhelming majority of introductory integration exercises, and each step shows the exponent being raised and the division being done rather than jumping to the result.

It does not yet integrate trig, exponential or logarithmic functions, and it does not do substitution or integration by parts. Those are reported honestly rather than approximated. The derivative side of the engine is considerably broader, so if you need to check an antiderivative you found by hand, differentiate it in the derivative calculator and confirm you get the original back. That check is valid for every method, including the ones this page cannot perform.

Why the check always works

Differentiation is the reliable direction

Differentiation is mechanical: there is always a rule that applies and it always terminates. Integration is not; it is closer to pattern recognition, and plenty of perfectly ordinary functions have no elementary antiderivative at all.

That asymmetry has a practical consequence worth internalising. Verifying an integral is easy even when finding it was hard, because verification only requires the easy direction. Whatever method produced your antiderivative, differentiating it back is a complete proof that it is right.

Common questions

Integral calculator FAQ

How do you integrate a polynomial?

Term by term, using the reverse power rule: add one to the exponent and divide by the new exponent. So ∫3x² dx = 3x³/3 = x³. A constant like 5 is 5x⁰, so it integrates to 5x. Add + C at the end for the constant of integration.

Why do you add + C to an integral?

Because differentiating destroys constants. x³/3, x³/3 + 7 and x³/3 − 1000 all have the same derivative, so reversing the process cannot tell which constant was there. The + C stands for the whole family of functions that share that derivative, and leaving it out costs marks.

What is the integral of 1/x?

ln|x| + C. The power rule fails here because 1/x is x⁻¹, and adding one to the exponent gives zero, which you cannot divide by. The absolute value matters, because 1/x is defined for negative x while ln alone is not.

Do definite integrals need a constant of integration?

No. A definite integral evaluates the antiderivative at both limits and subtracts, so the constant appears twice and cancels. Only indefinite integrals carry + C.

How do you check an integral is correct?

Differentiate your answer. If it returns the original expression exactly, the integral is right; if not, something went wrong. This check works no matter which method produced the answer, and it takes seconds.

What is the difference between a definite and an indefinite integral?

An indefinite integral returns a family of functions and carries + C. A definite integral has limits, returns a single number, and equals F(b) − F(a) where F is any antiderivative. Geometrically the definite integral is the signed area between the curve and the x-axis.

Can this integral calculator do substitution or integration by parts?

Not yet. It handles polynomial integration exactly, with steps, and says so plainly for anything else rather than guessing. For those methods, work the substitution yourself and verify the result by differentiating it in the derivative calculator, which is a complete check.

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