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AP® Calculus BC · Unit 6 · Topic 6.12

Integrating Using Linear Partial Fractions: Revision Notes (Calculus BC 6.12)

One-page recap of linear partial fractions: when the method applies, two ways to find the constants, integrating each piece as a logarithm, and the mistakes that cost marks.

Short on time? This page is the recap. For the full method, the graph and worked examples, use the full study guide. BC-only material. Prerequisites (FTC 6.7, ∫ 1/x dx 6.8, substitution 6.9, long division 6.10) are on the Calculus BC roadmap.

Recap

  • Partial fractions is adding fractions backwards: split one hard fraction into simple fractions with linear denominators.
  • Use it when the fraction is proper and the bottom factors into different linear factors.
  • Try substitution first. If the top is a multiple of the derivative of the bottom, you get a single ln straight away.
  • The course does not use repeated factors such as (x − 1)² or quadratic factors such as x² + 1.

Key relationships

Step What to write Note
Decompose N(x)/((x − p)(x − q)) = A/(x − p) + B/(x − q) Three factors: add C/(x − r)
Clear fractions N(x) = A(x − q) + B(x − p) True for every x
Find constants Put x = p, then x = q; or match coefficients Cover-up is the quick version
Integrate ∫ A/(x − p) dx = A ln|x − p| + C Keep the absolute values
Non-monic factor ∫ 1/(ax + b) dx = (1/a) ln|ax + b| + C Chain-rule factor 1/a
Improper fraction Divide first, then decompose the remainder Needed when top degree ≥ bottom degree
Example 3/(x(x + 3)) = 1/x − 1/(x + 3) x = 0 gives A = 1; x = −3 gives B = −1

Mistakes to avoid

  1. Log of the denominator. ∫ 1/(x² − 4) dx is not ln|x² − 4|; differentiate to see why.
  2. Wrong root. The factor (x + 3) is zero at x = −3.
  3. Missing 1/a for a factor like 2x − 1.
  4. Decomposing before dividing when the fraction is improper.
  5. Limits across an asymptote. If a root of the bottom lies between the limits, the integral is improper (Topic 6.13).
  6. No check. Recombine the fractions; you should get the original top.

Quick self-check

  1. Decompose 3/(x² + 3x). (1/x − 1/(x + 3))
  2. Find ∫ 1/((x − 1)(x − 2)) dx. (ln|x − 2| − ln|x − 1| + C, which is ln|(x − 2)/(x − 1)| + C)
  3. Evaluate ∫ from 0 to 1 of 2/((x + 1)(x + 3)) dx. (ln(3/2), about 0.405, from 1/(x + 1) − 1/(x + 3))

Next: practice questions.

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