Short on time? This page is the recap. For the decision path, the lookalike table and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Notation: ∫ f(x) dx is an indefinite integral (+ C); ∫ (a to b) f(x) dx is a definite integral.
Recap
- There is no new formula in Topic 6.14. The skill is choosing the technique by looking at the integrand.
- Ask in order: basic rule? rewrite? substitution? long division? complete the square? (BC: parts? partial fractions?)
- After a rewrite, division or substitution, the new integral often goes back to the basic rules.
- Some integrands, such as e^(−x²) and sin(x²), have no antiderivative in familiar functions. Use technology for a definite value, or write an accumulation function.
Key relationships
| Technique | Signal | Example |
|---|---|---|
| Basic rule | Matches a known derivative | ∫ sec²x dx = tan x + C |
| Rewrite | Expandable product, fraction over one term, roots | ∫ (x + 1)/x dx = ∫ (1 + 1/x) dx |
| Substitution | Inner function g(x) with g′(x) present, up to a constant | ∫ x/(x² + 9) dx = ½ ln(x² + 9) + C |
| Long division | Rational, degree of top ≥ degree of bottom | x²/(x² + 9) = 1 − 9/(x² + 9) |
| Complete the square | Quadratic below with no real roots, no matching x on top | x² − 6x + 13 = (x − 3)² + 4 |
| arctan form | 1/(u² + k²) | ∫ 1/(u² + k²) du = (1/k) arctan(u/k) + C |
| Parts (BC only) | Product of unlike types, e.g. power × log | ∫ u dv = uv − ∫ v du |
| Partial fractions (BC only) | Proper fraction, different linear factors below | 6/((x − 1)(x + 5)) = 1/(x − 1) − 1/(x + 5) |
Assumptions behind the methods
- Substitution needs the derivative of the inner function present up to a constant factor, never a leftover variable.
- For a definite integral by substitution, change the limits to u-values (or return to x and use the original limits).
- ln|u| needs absolute value bars unless u is always positive, as with x² + 9.
Mistakes to avoid
- Starting substitution with no signal. Check for g′(x) first.
- Writing ln(denominator) for every fraction. ∫ 1/(x² + 9) dx is an arctan, not a log.
- Skipping long division when the top’s degree is at least the bottom’s.
- Integrating top and bottom separately.
- Old limits after substitution in a definite integral.
- Assuming every integrand has a formula.
Quick self-check
- Find ∫ x/(x² + 1)² dx. (−1/(2(x² + 1)) + C, substitution with u = x² + 1)
- Find ∫ (x + 1)/x dx. (x + ln|x| + C, split the fraction first)
- Find ∫ 1/(x² + 2x + 5) dx. (½ arctan((x + 1)/2) + C, since x² + 2x + 5 = (x + 1)² + 4)
Next: practice questions.