AP Calculus AB and BC

Integral of x^n: The Power Rule for Antiderivatives

Q: What is the integral of x^n?

\int x^n\,dx = \frac{x^{n+1}}{n+1} + C for every real exponent n except n = -1. Raise the exponent by one and divide by the new exponent. For example, \int x^3\,dx = \frac{x^4}{4} + C, and differentiating that returns x^3.

Q: Why does the power rule for integrals exclude n = -1?

Because the formula divides by n + 1. At n = -1 that is division by zero, so the rule gives nothing. The correct antiderivative there is \int \frac{1}{x}\,dx = \ln|x| + C, which you can confirm by differentiating \ln|x|.

Q: Does the power rule work for negative and fractional exponents?

Yes, for every exponent except -1. For instance \int x^{-3}\,dx = -\frac{1}{2}x^{-2} + C and \int x^{1/2}\,dx = \frac{2}{3}x^{3/2} + C. Rewrite roots and quotients as powers first, then apply the rule.

Q: What is the integral of (ax + b)^n?

Substitute u = ax + b, so du = a\,dx. The result is \frac{(ax+b)^{n+1}}{a(n+1)} + C for n \neq -1. The extra \frac{1}{a} is what the plain power rule misses, which is why \int (3x+1)^4\,dx = \frac{(3x+1)^5}{15} + C.

The integral of x^n is x^(n+1)/(n+1) + C, valid for every real exponent n except n = -1. Raise the exponent by one, then divide by the new exponent. At n = -1 the rule would divide by zero, and the antiderivative of 1/x is ln|x| + C instead.

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

How the power rule for integrals works

To antidifferentiate a power of $x$ , raise the exponent by one and divide by that new exponent. This is the reverse of the derivative power rule, which multiplies by the exponent and drops it by one.

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

You do not have to trust the formula. Differentiate the answer and see whether the integrand comes back. The constant $\frac{1}{n+1}$ rides along, and the derivative power rule brings down a factor of $n+1$ that cancels it.

\frac{d}{dx}\left[\frac{x^{n+1}}{n+1}\right] = \frac{1}{n+1} \cdot (n+1)x^{n} = x^{n}

That cancellation is the whole proof, and it is also why the rule breaks exactly once. The step is legal only when $n + 1 \neq 0$ .

Check every antiderivative by differentiating

Antidifferentiation has a free verification step that differentiation does not: take the derivative of your answer and compare it with the integrand. On the AP exam this costs about ten seconds and catches a dropped coefficient or a mis-shifted exponent before it costs you the point.

Why the rule fails at n = -1

Substituting $n = -1$ into the formula gives $\frac{x^{0}}{0}$ , and dividing by zero is undefined. So the power rule says nothing at all about $\int x^{-1}\,dx$ , and that integral needs a different antiderivative.

\int \frac{1}{x}\,dx = \ln|x| + C

The absolute value matters. $\ln x$ alone is defined only for $x > 0$ , but $\frac{1}{x}$ exists on both sides of the origin, and $\ln|x|$ has derivative $\frac{1}{x}$ for every $x \neq 0$ . Note that no antiderivative bridges $x = 0$ , since $\frac{1}{x}$ is not continuous there, so the constant $C$ can differ on the two intervals.

One exception, not a family of them

Every other exponent is fine, including negative ones and fractions. $\int x^{-2}\,dx = -x^{-1} + C$ and $\int \sqrt{x}\,dx = \int x^{1/2}\,dx = \frac{2}{3}x^{3/2} + C$ both come straight from the power rule. Only $n = -1$ is special.

Where the power rule for integrals shows up on the AP exam

The rule is introduced in Unit 6, Topic 6.8 (Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation). Unit 6 carries a weighting of 15 to 20 percent on both AB and BC, the largest single unit on either exam.

It rarely appears bare. More often it is the last step after you have rewritten the integrand into powers of $x$ , or the step you take once a substitution has cleared the way.

Topic 6.7, evaluating definite integrals: find the antiderivative with the power rule, then apply the Fundamental Theorem of Calculus.
Topic 6.9, substitution: after a $u$ -substitution the integral usually collapses to $\int u^n\,du$ , which is this rule in a different letter.
Topic 6.10, long division and completing the square: division turns an improper rational function into a polynomial plus a remainder, and the polynomial part integrates term by term with the power rule.
Topic 6.14, selecting techniques: recognizing that an integrand is already a sum of powers is what tells you no technique is needed.

Rewriting is the skill being tested as much as the rule itself. Roots and quotients hide powers, so convert before you integrate.

\int \frac{x^3 + 2}{x^2}\,dx = \int \left(x + 2x^{-2}\right)dx = \frac{x^2}{2} - \frac{2}{x} + C

Common mistakes with the integral of x^n

Applying the derivative rule by reflex. Antidifferentiating $x^4$ gives $\frac{x^5}{5}$ , not $4x^3$ . Raise the exponent, do not lower it.
Forgetting to divide. Writing $\int x^5\,dx = x^6 + C$ leaves out the $\frac{1}{6}$ . Differentiating your answer catches this instantly.
Using the power rule on $\frac{1}{x}$ . It produces $\frac{x^0}{0}$ , which is undefined. The answer is $\ln|x| + C$ .
Dropping the absolute value in $\ln|x|$ , which quietly narrows the answer to $x > 0$ .
Omitting $+ C$ on an indefinite integral. The antiderivative is a whole family of functions, and the missing constant is a standard scoring deduction.
Treating a composite as a simple power. $\int (3x+1)^4\,dx$ is not $\frac{(3x+1)^5}{5}$ , because the chain rule leaves an extra factor of 3. Substitute $u = 3x+1$ and divide by 3.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you. The page states the general rule for every exponent n except -1, but a numeric check needs a concrete exponent, so it fixes n = 3: it verifies that the derivative of x^4/4 is x^3, which is the rule with n + 1 = 4.

Frequently asked questions

What is the integral of x^n?

$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ for every real exponent $n$ except $n = -1$ . Raise the exponent by one and divide by the new exponent. For example, $\int x^3\,dx = \frac{x^4}{4} + C$ , and differentiating that returns $x^3$ .

Why does the power rule for integrals exclude n = -1?

Because the formula divides by $n + 1$ . At $n = -1$ that is division by zero, so the rule gives nothing. The correct antiderivative there is $\int \frac{1}{x}\,dx = \ln|x| + C$ , which you can confirm by differentiating $\ln|x|$ .

Does the power rule work for negative and fractional exponents?

Yes, for every exponent except $-1$ . For instance $\int x^{-3}\,dx = -\frac{1}{2}x^{-2} + C$ and $\int x^{1/2}\,dx = \frac{2}{3}x^{3/2} + C$ . Rewrite roots and quotients as powers first, then apply the rule.

What is the integral of (ax + b)^n?

Substitute $u = ax + b$ , so $du = a\,dx$ . The result is $\frac{(ax+b)^{n+1}}{a(n+1)} + C$ for $n \neq -1$ . The extra $\frac{1}{a}$ is what the plain power rule misses, which is why $\int (3x+1)^4\,dx = \frac{(3x+1)^5}{15} + C$ .