Power rule for integrals: Integrate x^n when n is not -1

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | FAQ

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The Principle

The move: Add $1$ to the exponent and divide by that new exponent, then append $+C$.

The invariant: This produces an antiderivative family whose derivative returns the original integrand $x^n$, provided the new denominator $n+1$ is not zero.

Pattern: $\int x^n\,dx \quad \longrightarrow \quad \frac{x^{n+1}}{n+1}+C$

Legal ✓	Illegal ✗
$\int x^4\,dx \to \frac{x^5}{5}+C$	$\int x^{-1}\,dx \not\to \frac{x^0}{0}+C$

Left: $n=4$, so $n+1=5 \neq 0$ and the rule applies. Right: $n=-1$, so the denominator becomes $0$; the expression looks eligible, but the condition fails and the move is not available.

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Conditions of Applicability

Condition: $n \neq -1$

Before applying, check: confirm the integrand is exactly of the form $x^n$ with a constant exponent, then verify that adding $1$ to the exponent does not produce $0$ in the denominator.

If the condition is violated: the formula creates division by zero, so it does not produce a valid antiderivative. The $n=-1$ case must use $\int \frac{1}{x}\,dx = \ln\left|x\right| + C$ instead.

• The rule covers positive exponents, zero, and negative exponents other than $-1$.
• The near-miss is $x^{-1}$: it has the right surface form, but the condition fails exactly where the denominator $n+1$ would become zero.
• If the integrand is not a pure power of $x$, pause and identify the actual local pattern before moving. The $x^{-1}$ exception belongs to a logarithm antiderivative, not to this power-rule family.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Failure Modes

Failure mode: apply the power rule mechanically to $x^{-1}$ and write $\frac{x^0}{0}+C$ -> produces a meaningless denominator and hides that this is the special logarithm case.

Debug: before integrating, compute the new exponent denominator $n+1$. If it equals $0$, stop and switch to the logarithm antiderivative for $\frac{1}{x}$.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• Why does adding $1$ to the exponent and then dividing by that same new exponent undo differentiation of $x^{n+1}$?
• Why is the condition $n \neq -1$ attached to the denominator rather than the original exponent alone?

For the Principle

• What quick check can you run before applying the rule to decide whether the $x^{-1}$ exception is present?
• How does the result change when $n$ is negative but not $-1$, such as $n=-3$?

Between Principles

• The derivative power rule and the power rule for integrals reverse each other. What does each rule do to the exponent, and where does the exception appear only on the integral side?

Generate an Example

• Build one integral where the power rule works and one near-miss where it fails, then explain which condition check separates them.

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the move in one sentence: _____Add 1 to the exponent, divide by the new exponent, and add C to integrate x^n when n is not -1.

Write the canonical equation: _____$\int x^n\,dx = \frac{x^{n+1}}{n+1}+C$

State the canonical condition: _____$n \neq -1$

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Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to use worked examples effectively.)

Procedure Walkthrough

Starting from $\int x^{-3}\,dx$, reach a simplified antiderivative.

Step	Expression	Operation
0	$\int x^{-3}\,dx$	-
1	$\frac{x^{-3+1}}{-3+1}+C$	Power rule - $n=-3$, so $n+1=-2 \neq 0$
2	$\frac{x^{-2}}{-2}+C$	Arithmetic on the exponent and denominator
3	$-\frac{1}{2}x^{-2}+C$	Simplify the coefficient
4	$-\frac{1}{2x^2}+C$	Rewrite the negative exponent as a fraction

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Drills

Forward step (Format A)

Apply the power rule once.

$\int x^6\,dx$

Reveal

$n=6$, so $n+1=7 \neq 0$.

$\int x^6\,dx = \frac{x^7}{7}+C$

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Apply the power rule once.

$\int x^{-3}\,dx$

Reveal

$n=-3$, so $n+1=-2 \neq 0$.

$\int x^{-3}\,dx = \frac{x^{-2}}{-2}+C = -\frac{1}{2x^2}+C$

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Can the power rule be applied? Explain the condition check before doing anything else.

$\int x^{-1}\,dx$

Reveal

No. Here $n=-1$, so $n+1=0$. That breaks the denominator in the rule, so the power rule is not applicable.

This is the required near-miss: the integrand has the form $x^n$, but the condition fails exactly at the exception. Use the integral of $\frac{1}{x}$ instead:

$\int x^{-1}\,dx = \int \frac{1}{x}\,dx = \ln\left|x\right| + C$

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Which integrands can be handled directly by the power rule for integrals?

(i) $x^5$ \quad (ii) $x^{-1}$ \quad (iii) $x^{-2}$ \quad (iv) $e^x$

Reveal

(i) and (iii) only.

• $x^5$: $n=5$, so $n+1=6 \neq 0$ -> valid
• $x^{-1}$: $n=-1$, so $n+1=0$ -> invalid
• $x^{-2}$: $n=-2$, so $n+1=-1 \neq 0$ -> valid
• $e^x$: not of the form $x^n$ -> use the exponential integral rule instead

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Action label (Format B)

What was done between these two steps, and was it valid?

$\int x^8\,dx \quad \longrightarrow \quad \frac{x^9}{9}+C$

Reveal

Power rule for integrals applied. The exponent was increased from $8$ to $9$, then divided by the new exponent. This is valid because $8+1=9 \neq 0$.

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What was done between these two steps? Was the move valid?

$\int x^{-1}\,dx \quad \longrightarrow \quad \frac{x^0}{0}+C$

Reveal

An invalid use of the power rule was attempted. The move is not valid because $n=-1$, so the new denominator becomes $0$.

The correct action is to reject the move and switch to:

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

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What was done between these two steps, and why is the move legal?

$\int x^{-4}\,dx \quad \longrightarrow \quad -\frac{1}{3}x^{-3}+C$

Reveal

Power rule for integrals applied, then simplified. Since $n=-4$, the new exponent is $-3$ and the new denominator is $-3$, which is nonzero:

$\int x^{-4}\,dx = \frac{x^{-3}}{-3}+C = -\frac{1}{3}x^{-3}+C$

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Transition identification (Format C)

Which transition uses the power rule for integrals?

$\int x^3\,dx \xrightarrow{(1)} \frac{x^4}{4}+C \xrightarrow{(2)} \frac{1}{4}x^4 + C$

Reveal

Transition (1) uses the power rule for integrals. Transition (2) is only algebraic rewriting of the coefficient.

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Which transition is invalid, and why?

$\int x^{-1}\,dx \xrightarrow{(1)} \frac{x^0}{0}+C \xrightarrow{(2)} \text{undefined}$

Reveal

Transition (1) is invalid. The power-rule pattern was attempted where $n=-1$, so the denominator becomes $0$ immediately.

The fix is not to simplify further. The fix is to reject the move and replace the chain with:

$\int x^{-1}\,dx = \ln\left|x\right| + C$

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Goal micro-chain (Format D)

Starting from $\int (x^2 + x^{-1})\,dx$, reach a correct final antiderivative by using the power rule only where it is legal.

Reveal

First use the integral sum rule to split the sum:

$\int (x^2 + x^{-1})\,dx = \int x^2\,dx + \int x^{-1}\,dx$

Apply the power rule to the first term only:

$\int x^2\,dx = \frac{x^3}{3}+C_1$

Use the integral of $\frac{1}{x}$ on the exception term:

$\int x^{-1}\,dx = \ln\left|x\right| + C_2$

Combine constants:

$\int (x^2 + x^{-1})\,dx = \frac{x^3}{3} + \ln\left|x\right| + C$

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Integrate $x^{-4}$ and simplify the final antiderivative.

Full solution

Step	Expression	Move
0	$\int x^{-4}\,dx$	-
1	$\frac{x^{-4+1}}{-4+1}+C$	Power rule - $n=-4$, so $n+1=-3 \neq 0$
2	$\frac{x^{-3}}{-3}+C$	Arithmetic on the exponent and denominator
3	$-\frac{1}{3}x^{-3}+C$	Simplify the coefficient
4	$-\frac{1}{3x^3}+C$	Rewrite the negative exponent as a fraction

Check: differentiating $-\frac{1}{3x^3}+C$ returns $x^{-4}$.

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Related Principles

Principle	Relationship
Indefinite integral as antiderivative	Explains what an antiderivative family means and why every result ends with $+C$
Derivative power rule	The inverse move: differentiating $x^{n+1}$ gives the integrand that the integral rule reverses
Derivative of ln(x)	Shows why the exceptional antiderivative family is logarithmic rather than another power

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FAQ

What is the power rule for integrals?

The power rule for integrals states that $\int x^n\,dx = \frac{x^{n+1}}{n+1}+C$ when $n \neq -1$. You add $1$ to the exponent, divide by the new exponent, and append the constant of integration.

Why does the rule exclude n = -1?

If $n=-1$, then the denominator becomes $n+1=0$, so the formula breaks immediately. That case is not a minor algebra issue; it marks a different antiderivative family: $\int \frac{1}{x}\,dx = \ln\left|x\right| + C$.

Does the rule work for negative exponents?

Yes, as long as the exponent is not $-1$. For example, $\int x^{-3}\,dx = -\frac{1}{2x^2}+C$ is valid because the new denominator is $-2$, not $0$.

Why do I still add +C after using the formula?

Indefinite integrals describe a family of antiderivatives, not one single function. The $+C$ keeps every vertical shift of the antiderivative in the answer, which is required for the full family.

How do I know whether to use the power rule or the integral of 1/x?

Check the exponent first. If the integrand is $x^n$ with $n \neq -1$, use the power rule. If the exponent is exactly $-1$, rewrite the integrand as $\frac{1}{x}$ and use the integral of $\frac{1}{x}$ instead.

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How This Fits in Unisium

In Unisium, integration fluency is trained as move selection plus guard checking, not as blind formula recall. For the power rule, the critical habit is to test the exponent before you act: most powers of $x$ take the standard antiderivative, but the single near-miss $x^{-1}$ changes the rule family completely. The drills above train that branch point directly, while retrieval practice, self-explanation, and elaborative encoding make the condition and exception stable enough to use under time pressure.

Explore further:

• Indefinite integral as antiderivative - See why every correct result ends with $+C$
• Integral of 1/x - Compare the standard power-family move with the single reciprocal exception at $n=-1$
• Derivative of ln(x) - See why the exception produces a logarithm rather than another power
• Derivative power rule - Connect the integration rule to its inverse derivative move

Ready to master the power rule for integrals? Start practicing with Unisium or explore the full learning framework in Masterful Learning.

Masterful Learning

The study system for physics, math, & programming that works: retrieval, connection, explanation, problem solving, and more.

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