← Calculus II

Unit 6: Partial Fractions

Splitting a tangled rational function into pieces simple enough to integrate one at a time.

5 topics · 21 questions

Each topic is a path, not a menu: work down it in order and every question builds on the one above. Click any of them to watch the answer draw itself.

Splitting a Rational Function into Simple Pieces

  1. How it worksHow does clearing denominators and matching coefficients recover the two hidden fractions inside a rational function?
  2. Worked exampleFind the partial fraction decomposition of (3x + 5)/((x + 1)(x + 2)).
  3. Common mistakeWhy doesn't a rational function need partial fractions if you can already integrate its denominator directly?
  4. Ties togetherHow does splitting a fraction into partial fractions reverse combining fractions over a common denominator?
Your turn1 of 2
Find F(x), the antiderivative of f(x) = (4x + 4)/((x + 1)(x + 2)), satisfying F(0) = 0.

Distinct Linear Factors

  1. How it worksHow does plugging in the root of each linear factor solve for that factor's constant almost instantly?
  2. Worked exampleFind the integral of (7x - 1)/((x - 1)(x + 3)) dx using partial fractions.
  3. Common mistakeWhy doesn't a repeated root show up as just one fraction the way a distinct root does?
  4. Worked exampleFind the integral of (2x + 3)/(x2 - x - 6) dx by factoring the denominator first.
  5. Going furtherWhat happens to the setup when three distinct linear factors appear in the denominator instead of two?
Your turn1 of 2
Find F(x), the antiderivative of f(x) = (4x + 4)/((x - 1)(x + 3)), satisfying F(2) = 0.

Repeated Linear Factors

  1. How it worksHow does each power of a repeated factor, from 1 up to its full exponent, get its own numerator constant?
  2. Going furtherWhat happens to the decomposition when a repeated factor and a distinct factor share one denominator?
  3. Worked exampleFind the partial fraction decomposition of (5x2 - 3)/((x - 2)3).
  4. Common mistakeWhy isn't (x - 2)3 in the denominator handled the same way as three distinct linear factors?
Your turn1 of 2
Find F(x), the antiderivative of f(x) = (3x2 + 4)/(x - 2)3, satisfying F(3) = 0.

Irreducible Quadratic Factors

  1. How it worksHow does an irreducible quadratic like x2 + 4 in the denominator pair with a numerator of the form Ax + B?
  2. Worked exampleFind the integral of (x + 1)/(x(x2 + 4)) dx using partial fractions.
  3. Common mistakeWhy doesn't every quadratic in a denominator need the Ax + B treatment?
  4. Worked exampleFind the integral of (3x2 + 2)/((x2 + 1)(x - 1)) dx using partial fractions.
Your turn1 of 2
Find F(x), the antiderivative of f(x) = (3x + 3)/(x(x2 + 6)), satisfying F(1) = 0.

Improper Rational Functions and Long Division First

  1. How it worksHow does polynomial long division peel off a polynomial part, leaving a proper fraction to decompose?
  2. Going furtherWhat happens to the long-division remainder once the numerator's degree drops below the denominator's?
  3. Worked exampleFind the integral of (x3 + 1)/(x2 - 1) dx by dividing first.
  4. Common mistakeWhy doesn't applying partial fractions directly to an improper rational function just work anyway?
Your turn1 of 2
Find F(x), the antiderivative of f(x) = (x3 + 4)/(x2 - 1), satisfying F(2) = 0.