Power Rule for Integrals Explained — Derivation, Examples & Quizzes

Q: Why is n = −1 excluded from the power rule?

Plugging n = −1 gives x⁰/0, which is division by zero and meaningless. The missing case is ∫ dx/x = ln|x| + C, a genuinely different antiderivative.

Q: How do I integrate √x or 1/x² with the power rule?

Rewrite them as powers first: √x = x^(1/2) gives (2/3)x^(3/2) + C, and 1/x² = x^(−2) gives −1/x + C. The rule only sees exponents, so put everything in exponent form.

Where it comes from

The problem: differentiation has a power rule, d/dx [xⁿ] = n·xⁿ⁻¹. Integration asks the reverse question — which function has derivative xⁿ? That is, find F with F′ = xⁿ.

The tempting guess just bumps the exponent:

Before reading on: the derivative power rule turns x³ into 3x². If you had to run that process backwards to integrate x², what would you do to undo the ×3? Commit to a guess before the next paragraph reveals the rule.

\int x n dx = x n+1 + C ?? the tempting — and wrong — guess

Kill it with n = 2. The guess claims ∫ x² dx = x³ + C. But differentiate the answer to check:

d/dx [x³ + C]

=

3x²

Not x² — the guess is three times too big. The derivative power rule multiplied by (n+1) = 3 on the way down, and the guess never undid it.

F(x)

=

x³/3

Divide by that extra factor: d/dx [x³/3] = 3x²/3 = x². That division is the power rule for integrals.

The same repair works for every exponent: xⁿ⁺¹ differentiates to (n+1)·xⁿ, so dividing by (n+1) leaves exactly xⁿ.

Two loose ends. First, the +C: since d/dx [C] = 0, adding any constant to F still gives F′ = xⁿ. There are infinitely many antiderivatives, and +C records the whole family in one symbol. Second, the n ≠ −1: plugging n = −1 gives x&sup0;/0 — division by zero, meaningless. That one missing exponent gets its own formula, ∫ dx/x = ln|x| + C (the Reciprocal entry on the sheet).

Derivation

We verify the candidate F(x) = xⁿ⁺¹/(n+1) for n ≠ −1. The strategy is the oldest one in the book: guess, then differentiate to check. If the derivative of our guess is xⁿ, the guess is an antiderivative.

F(x)

=

xⁿ⁺¹/(n+1) = (1/(n+1)) · xⁿ⁺¹

Step 1 — pull out the constant. 1/(n+1) does not depend on x, so it factors out of the derivative (constant-multiple rule).

F′(x)

=

(1/(n+1)) · d/dx [xⁿ⁺¹]

Step 2 — set up the derivative power rule. The exponent n+1 is just a number, so the ordinary power rule applies.

=

(1/(n+1)) · (n+1) · xⁿ⁺¹⁻¹

Step 3 — apply it. d/dx [xⁿ⁺¹] = (n+1)·xⁿ. The (n+1) that broke the naive guess reappears here — on purpose.

=

xⁿ

Step 4 — cancel. (1/(n+1))·(n+1) = 1 (legal because n ≠ −1). So F′ = xⁿ: F is an antiderivative. ∎

Why is +C the whole story? If G′ = xⁿ too, then (G−F)′ = 0, and a function with zero derivative everywhere is constant (a corollary of the Mean Value Theorem). So G = F + C for some constant C — every antiderivative is captured.

How to use it

Before reading on: the procedure below has five steps, but one of them is the step everyone skips. Which step do you think it is — and what goes wrong in the check if you skip it?

The procedure, every time:

Write everything as xⁿ. The rule only sees exponents: √x = x^1/2, 1/x² = x⁻², x·√x = x^3/2. Do this rewrite first.
Add 1 to the exponent. n becomes n+1.
Divide by the new exponent. This is the step everyone skips — the division undoes the (n+1) the derivative would multiply by.
Add +C — once, at the very end, not once per term.
Check by differentiating. Five seconds: does your answer differentiate back to the integrand? If yes, you are done.

Power rule or something else?

Sums: integrate term by term — ∫ (4x³ − 2x + 5) dx applies the rule three times. 1/x: that is n = −1 — forbidden here, use ∫ dx/x = ln|x| + C. Products like x²·sin x: there is no product rule for integrals — this needs other techniques.

Tricky exponents

\int 1 dx = \int x 0 dx = x 1 /1 + C = x + C n = 0 still works — the rule never skips Say it: the integral of one d x equals x plus C — the power rule works for n equals zero too

Negative exponents are fine (except −1):

∫ x⁻³ dx

=

x⁻²/(−2) + C

Add one to the exponent, divide by the new exponent.

=

−1/(2x²) + C

Tidy the fraction.

Fractional exponents are fine: dividing by 3/2 is the same as multiplying by 2/3.

Common mistake: writing ∫ x³ dx = x⁴ + C and stopping — forgetting to divide by the new exponent. Differentiate to check: 4x³ ≠ x³. The division is not optional.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Example 1 — the basic move: ∫ x⁵ dx

Add 1 to the exponent. 5 becomes 6: x⁶.
Divide by the new exponent. x⁶/6.
Add +C. = x⁶/6 + C.
Check by differentiating. d/dx [x⁶/6 + C] = 6x⁵/6 = x⁵. Matches ✓

Common mistake: answering x⁶ + C (no division). The check in step 4 catches it: d/dx [x⁶] = 6x⁵ ≠ x⁵.

Your turn: Compute ∫ x⁷ dx.

Answer: x⁸/8 + C

Add 1 to the exponent: 7 becomes 8. Divide by the new exponent: x⁸/8. Check: d/dx [x⁸/8] = 8x⁷/8 = x⁷. ✓

Example 2 — a radical: ∫ √x dx

Rewrite as a power first. √x = x^1/2. (Why? The rule only sees exponents.)
Add 1: 1/2 + 1 = 3/2. Divide by 3/2 — i.e. multiply by 2/3: (2/3)x^3/2.
Add +C: (2/3)x^3/2 + C.
Check. d/dx [(2/3)x^3/2] = (2/3)(3/2)x^1/2 = x^1/2 = √x. Matches ✓

Common mistake: dividing by 1/2 (the old exponent) instead of 3/2 (the new one). Always divide by the exponent after adding one.

Your turn: Compute ∫ ∛x dx (the cube root).

Answer: (3/4)x^4/3 + C

Rewrite first: ∛x = x^1/3. Add 1: 1/3 + 1 = 4/3. Divide by 4/3 (multiply by 3/4): (3/4)x^4/3. Check: d/dx [(3/4)x^4/3] = (3/4)(4/3)x^1/3 = x^1/3. ✓

Example 3 — term by term: ∫ (4x³ − 2x + 5) dx

Split into three power-rule integrals. ∫4x³ dx − ∫2x dx + ∫5 dx. (Why allowed? The sum rule for integrals.)
Constants factor out. 4·∫x³ dx − 2·∫x dx + 5·∫1 dx.
Apply the rule to each. 4·(x⁴/4) − 2·(x²/2) + 5x = x⁴ − x² + 5x + C. (Note: ∫5 dx = ∫5x⁰ dx = 5x — the n = 0 case.)
Check term by term. d/dx gives 4x³ − 2x + 5. Matches ✓ (One +C at the end covers all three terms.)

Common mistake: writing +C after every term (C₁ + C₂ + C₃). A sum of arbitrary constants is still one arbitrary constant — write +C once.

Your turn: Compute ∫ (3x⁴ + 6x² − 2) dx.

Answer: (3/5)x⁵ + 2x³ − 2x + C

Term by term: 3·(x⁵/5) + 6·(x³/3) − 2x, with one +C at the end. Check: d/dx gives 3x⁴ + 6x² − 2. ✓

Example 4 — negative exponent: ∫ (x² + 1/x²) dx

Rewrite. 1/x² = x⁻². The exponent is −2 — not −1, so the power rule is legal.
First term: ∫x² dx = x³/3.
Second term: ∫x⁻² dx = x⁻¹/(−1) = −x⁻¹ = −1/x.
Combine: x³/3 − 1/x + C.
Check. d/dx [−1/x] = d/dx [−x⁻¹] = x⁻² = 1/x². Matches ✓

Common mistake: seeing 1/x² and reaching for ln — but ln belongs to 1/x (exponent −1), not 1/x². Only the exact exponent −1 is forbidden; −2 is perfectly fine.

Your turn: Compute ∫ (x³ + 1/x³) dx.

Answer: x⁴/4 − 1/(2x²) + C

Rewrite: 1/x³ = x⁻³ (not −1, so the power rule is legal). ∫x³ dx = x⁴/4; ∫x⁻³ dx = x⁻²/(−2) = −1/(2x²). Check: d/dx [−1/(2x²)] = x⁻³ = 1/x³. ✓

Memorization tips

Say it aloud: “add one, divide by the new one, plus C.” The rhythm matches the three moves in order.
The 5-second check: differentiate your answer. If it does not give back the integrand, something is wrong — this single habit catches most errors.
Anchor on n = 0: ∫ 1 dx = x + C. If you ever wonder whether the rule “skips” an exponent, this anchor says no.
Pair the exception: n = −1 → ln|x| + C. Drill them as a pair — “every power except negative one; negative one is log.”
Fractions: dividing by 3/2 is multiplying by 2/3. Do the flip confidently instead of freezing.
+C once: one constant at the very end, however many terms you integrated.

Final challenge

Five mixed questions — basics, traps, and one definite integral. Score 5/5 and the power rule is yours.

How to learn a formula here

Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
Work the examples with the answers covered, then uncover one step at a time and compare.
Finish with the final challenge — five mixed questions including the classic traps.
Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.

Frequently asked questions

What is the power rule for integrals?

The power rule for integrals says ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, for n ≠ −1: add one to the exponent, divide by the new exponent, and add the constant of integration C.

Why is n = −1 excluded from the power rule?

Plugging n = −1 gives x&sup0;/0, which is division by zero and meaningless. The missing case is ∫ dx/x = ln|x| + C, a genuinely different antiderivative.

Why do we add +C to an indefinite integral?

Derivatives kill constants, so xⁿ⁺¹/(n+1) + C differentiates to xⁿ for every constant C. The +C records the whole infinite family of antiderivatives at once.

How do I integrate √x or 1/x² with the power rule?

Rewrite them as powers first: √x = x^1/2 gives (2/3)x^3/2 + C, and 1/x² = x⁻² gives −1/x + C. The rule only sees exponents, so put everything in exponent form.

Does the power rule work on products like x²·sin x?

No — there is no product rule for integrals. The power rule handles single powers (and sums of them, term by term); a product like x²·sin x needs other techniques.

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