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AP® Calculus AB · Unit 6 · Topic 6.8

Antiderivatives and Indefinite Integrals: Basic Rules and Notation: Revision Notes (Calculus AB 6.8)

One-page recap of indefinite integrals: the meaning of + C, the table of basic antiderivatives, rewriting before integrating and the errors that cost marks.

Short on time? This page is the recap. For explanations, the family-of-curves figure and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.

Recap

  • ∫ f(x) dx (no limits) is the indefinite integral: all antiderivatives of f.
  • If F′ = f, then ∫ f(x) dx = F(x) + C. Antiderivatives on an interval differ only by a constant, so this lists them all.
  • Graphically, the family is one curve shifted up and down. All members have the same slope at each x.
  • Constant multiples and sums integrate term by term. Use one + C at the end.
  • No product or quotient rule: expand, split or use an identity first.
  • Some functions, such as e^(x²) and sin(x²), have no antiderivative you can write as a formula.

Key relationships

Integral Result
∫ xⁿ dx, n ≠ −1 xⁿ⁺¹/(n + 1) + C
∫ (1/x) dx ln|x| + C
∫ eˣ dx; ∫ aˣ dx eˣ + C; aˣ/ln a + C
∫ cos x dx; ∫ sin x dx sin x + C; −cos x + C
∫ sec²x dx; ∫ csc²x dx tan x + C; −cot x + C
∫ sec x tan x dx; ∫ csc x cot x dx sec x + C; −csc x + C
∫ 1/√(1 − x²) dx; ∫ 1/(1 + x²) dx arcsin x + C; arctan x + C

Assumptions behind the rules

  • Each result holds on an interval where the integrand is defined. For example, ln|x| + C works on x > 0 or on x < 0, not across 0.
  • The constant C is arbitrary. A known point on the graph (Topic 7.7) is needed to fix it. In a definite integral it cancels, so it does not matter there.

Mistakes to avoid

  1. No + C on an indefinite integral.
  2. Power rule on 1/x, which divides by 0.
  3. ln for every fraction: ∫ 1/x² dx = −1/x + C, not ln(x²).
  4. Factor-by-factor integration of a product.
  5. Trig sign slips: ∫ sin x dx = −cos x + C; ∫ csc²x dx = −cot x + C.
  6. No dx, or an “=” between a function and its integral.

Quick self-check

  1. Find ∫ x^(2/3) dx. ((3/5)x^(5/3) + C)
  2. Find ∫ 7 dx. (7x + C)
  3. Find ∫ csc²x dx. (−cot x + C, because d/dx cot x = −csc²x)
  4. Is ∫ (x + 1)² dx = (x + 1)³/3 + C correct? (Yes: its derivative is (x + 1)². Expanding to x² + 2x + 1 and integrating gives x³/3 + x² + x + C, which differs only by the constant 1/3.)

Next: practice questions.

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