What's the difference between the Power Rule for derivatives and Power Rule for Integration?

They are inverse operations. For derivatives, \frac{d}{dx}[x^n] = nx^{n-1} — you multiply by the old exponent and decrease it. For integration, \int x^n dx = \frac{x^{n+1}}{n+1} + C — you increase the exponent and divide by the new exponent. One reverses the other.

Why is the constant of integration C necessary in indefinite integrals?

Because the derivative of any constant is zero. For example, \frac{d}{dx}[x^2] = 2x and \frac{d}{dx}[x^2 + 7] = 2x and \frac{d}{dx}[x^2 - 100] = 2x all give the same derivative. So when you reverse the process and integrate 2x, you get x^2 plus any constant — that's why we write + C.

Why doesn't Power Rule work when n = -1?

If n = -1, then the denominator becomes n + 1 = 0, and you would be dividing by zero: \frac{x^0}{0}, which is undefined. The antiderivative of x^{-1} = \frac{1}{x} is \ln|x| + C, which is completely different from a power function. This is why n \neq -1 is a required restriction.

Can I use Power Rule if the exponent is a fraction or decimal?

Yes. Power Rule works for any real number n except -1, including fractions and decimals. For example, \int x^{0.5} dx = \frac{x^{1.5}}{1.5} + C and \int x^{-2/3} dx = \frac{x^{1/3}}{1/3} + C = 3x^{1/3} + C. Just add 1 to the exponent and divide by the result.

Power Rule for Integration

Power Rule for Integration instantly finds antiderivatives of polynomial terms; use it to integrate any power of x except x to the negative one.

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

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What each symbol means

What Power Rule for Integration takes

Symbol	Meaning
$n$	The exponent of $x$ ; it can be any real number except $-1$ because $n = -1$ would make the denominator $n + 1 = 0$ , which is undefined.
$x$	The variable being raised to the power $n$ ; in physics problems it often represents time or distance, and the integral finds a quantity whose rate of change is $x^n$ .
$C$	The constant of integration; it accounts for the fact that infinitely many antiderivatives exist (they differ only by a constant), so every indefinite integral includes $+ C$ to represent all of them.

When to use it

Use Power Rule whenever you encounter an integral of the form $x^n$ and need to find its antiderivative.

Level

Usually taught in: Calculus I · Appears on: AP Calculus

Worked examples

1. Integrate a cubic monomial

Problem

Find

\int x^3 \, dx

$n = 3$
We identify the exponent: this is $x^3$ .
$\frac{x^{3+1}}{3+1} + C$
We apply Power Rule: increase the exponent by 1 (from $3$ to $4$ ) and divide by the new exponent.
$\frac{x^4}{4} + C$
We simplify: $3 + 1 = 4$ , so the antiderivative is $\frac{x^4}{4}$ plus the constant of integration.

Answer: $\frac{x^4}{4} + C$

This is a straightforward application of Power Rule — we have a single power of $x$ , so we apply the rule directly. We can verify by differentiating: $\frac{d}{dx}\left(\frac{x^4}{4}\right) = \frac{4x^3}{4} = x^3$ , which is our original integrand.

2. Integrate a negative power

Problem

Find

\int x^{-2} \, dx

$n = -2$
The exponent is negative, but Power Rule still applies as long as $n \neq -1$ .
$\frac{x^{-2+1}}{-2+1}$
We apply Power Rule: add 1 to the exponent to get $-2 + 1 = -1$ , and divide by this new exponent.
$\frac{x^{-1}}{-1}$
We simplify the exponent in the numerator and the denominator.
$-\frac{1}{x} + C$
We rewrite $x^{-1}$ as $\frac{1}{x}$ and apply the negative sign in the denominator to get $-\frac{1}{x}$ .

Answer: $-\frac{1}{x} + C$

This example shows that Power Rule works with negative exponents, but we must avoid $n = -1$ , which would require the special formula $\int x^{-1} dx = \ln|x| + C$ instead. We can verify: $\frac{d}{dx}\left(-\frac{1}{x}\right) = \frac{d}{dx}(-x^{-1}) = -(-1)x^{-2} = x^{-2}$ .

3. Find a basketball player's score function from their scoring rate

Problem

A basketball player's scoring rate (points per week) during a season follows

\frac{dS}{dt} = 4t^{3/2}

, where

t

is measured in weeks since the start. Find the total score function

S(t)

if the player has 10 points at the end of week 1:

S(1) = 10

$S(t) = \int 4t^{3/2} \, dt$
Since scoring rate is the derivative of total score, we integrate to find the total score function.
$S(t) = 4 \int t^{3/2} \, dt$
We factor out the constant 4 from the integral.
$S(t) = 4 \cdot \frac{t^{3/2 + 1}}{3/2 + 1} + C$
We apply Power Rule with $n = \frac{3}{2}$ : add 1 to get $\frac{3}{2} + 1 = \frac{5}{2}$ , then divide by $\frac{5}{2}$ .
$S(t) = 4 \cdot \frac{t^{5/2}}{5/2} + C$
The new exponent is $\frac{5}{2}$ .
$S(t) = 4 \cdot \frac{2}{5} \cdot t^{5/2} + C$
We simplify the division by a fraction: $\frac{1}{5/2} = \frac{2}{5}$ .
$S(t) = \frac{8}{5}t^{5/2} + C$
We multiply: $4 \cdot \frac{2}{5} = \frac{8}{5}$ .
$10 = \frac{8}{5}(1)^{5/2} + C$
We apply the initial condition $S(1) = 10$ . Since $(1)^{5/2} = 1$ , this simplifies to $10 = \frac{8}{5} + C$ .
$C = 10 - \frac{8}{5} = \frac{50}{5} - \frac{8}{5} = \frac{42}{5}$
We solve for the constant: $C = 10 - \frac{8}{5} = \frac{50 - 8}{5} = \frac{42}{5}$ .

Answer: $S(t) = \frac{8}{5}t^{5/2} + \frac{42}{5}$

Power Rule works with fractional exponents too. By integrating the rate function and applying the initial condition, we found the complete score formula. The answer tells us the player's cumulative points at any time $t$ in the season.

Common mistakes

Where Power Rule for Integration usually goes wrong

The mistake	Why it is wrong	The fix
Writing $\int x^n dx = \frac{x^n}{n} + C$ , forgetting to add 1 to the exponent.	The exponent must increase by exactly 1; dividing by the original exponent $n$ gives the derivative, not the antiderivative.	Always remember: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ — you add 1 to the exponent and divide by the new exponent.
Attempting to use Power Rule when $n = -1$ , writing $\int x^{-1} dx = \frac{x^0}{0} + C$ or $\frac{1}{0} + C$ .	Division by zero is undefined. The Power Rule explicitly excludes $n = -1$ because $n + 1 = 0$ ; the correct antiderivative of $x^{-1} = \frac{1}{x}$ is $\ln\|x\|$ , not a power function.	Memorize the exception: $\int x^{-1} dx = \int \frac{1}{x} dx = \ln\|x\| + C$ , which requires a different approach entirely.
Forgetting to include the constant of integration: writing $\int (x^3 + 2x) dx = \frac{x^4}{4} + x^2$ without a $+ C$ .	An indefinite integral represents all antiderivatives; they differ by a constant. Without $+ C$ , you are claiming there is only one antiderivative, which is false.	Always end an indefinite integral with $+ C$ : $\int (x^3 + 2x) dx = \frac{x^4}{4} + x^2 + C$ .

Tips and when to use something else

Verify your answer by taking the derivative: if $\frac{d}{dx}\left[\frac{x^{n+1}}{n+1} + C\right] = x^n$ , you got it right.
Power Rule works for negative exponents, fractional exponents, and decimal exponents — any real number except $n = -1$ .
When you see $\int x^{-1} dx$ or $\int \frac{1}{x} dx$ , do not use Power Rule; instead use Integral of 1/x: $\ln|x| + C$ .
For integrals like $\int (3x^2 + x^{1/2}) dx$ , apply Power Rule separately to each term, then combine.

Related formulas

Indefinite IntegralThe indefinite integral finds all functions whose derivative equals a given function, expressed as an antiderivative plus an arbitrary constant.Integral of 1/xIntegrating 1/x yields the natural logarithm of the absolute value of x, plus a constant—essential for reciprocal function problems.U-SubstitutionU-substitution simplifies integrals by transforming them into easier forms using a change of variables, ideal when you spot a function and its derivative.Fundamental Theorem of CalculusCompute the exact value of a definite integral using the antiderivative—connect area under a curve to the accumulation of change.

Frequently asked questions

What's the difference between the Power Rule for derivatives and Power Rule for Integration?: They are inverse operations. For derivatives, $\frac{d}{dx}[x^n] = nx^{n-1}$ — you multiply by the old exponent and decrease it. For integration, $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ — you increase the exponent and divide by the new exponent. One reverses the other.
Why is the constant of integration $C$ necessary in indefinite integrals?: Because the derivative of any constant is zero. For example, $\frac{d}{dx}[x^2] = 2x$ and $\frac{d}{dx}[x^2 + 7] = 2x$ and $\frac{d}{dx}[x^2 - 100] = 2x$ all give the same derivative. So when you reverse the process and integrate $2x$ , you get $x^2$ plus any constant — that's why we write $+ C$ .
Why doesn't Power Rule work when $n = -1$ ?: If $n = -1$ , then the denominator becomes $n + 1 = 0$ , and you would be dividing by zero: $\frac{x^0}{0}$ , which is undefined. The antiderivative of $x^{-1} = \frac{1}{x}$ is $\ln|x| + C$ , which is completely different from a power function. This is why $n \neq -1$ is a required restriction.
Can I use Power Rule if the exponent is a fraction or decimal?: Yes. Power Rule works for any real number $n$ except $-1$ , including fractions and decimals. For example, $\int x^{0.5} dx = \frac{x^{1.5}}{1.5} + C$ and $\int x^{-2/3} dx = \frac{x^{1/3}}{1/3} + C = 3x^{1/3} + C$ . Just add 1 to the exponent and divide by the result.

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Reviewed 2026-09-18