Recipe

Integrate a rational function with repeated or irreducible quadratic factors

Use when: the denominator has a repeated linear factor (x-r)^k, an irreducible quadratic factor, or both, once distinct linear factors alone (recipe above) no longer cover every factor

Goal: every factor decomposed correctly, using extra terms for the repeated powers and a linear numerator for each irreducible quadratic, then integrated

Taught in Partial fractions — repeated and irreducible factors · extends Integrate a rational function with distinct linear factors

1. 1

   A repeated linear factor (x-r)^k contributes k separate terms, one for every power from 1 to k, each with its own unknown constant.

   Example

   \displaystyle\int\frac{2x+1}{(x-2)^2(x+1)}\,dx: set up \frac{2x+1}{(x-2)^2(x+1)} = \frac{A}{x-2}+\frac{B}{(x-2)^2}+\frac{C}{x+1}

   Why: a single term over (x-r)^k can't match every power of (x-r) that clearing denominators produces; only k separate terms can.

2. 2

   An irreducible quadratic factor (negative discriminant, no real roots) contributes a term with a linear numerator, \dfrac{Dx+E}{ax^2+bx+c}, not a constant one.

   Example

   If the denominator also had a factor x^2+1, that term would be set up as \dfrac{Dx+E}{x^2+1}, not \dfrac{D}{x^2+1}.

   Why: the numerator only needs degree one less than the quadratic denominator's degree 2 to keep the whole fraction proper, and a constant numerator would under-determine the system.

3. 3

   Cover-up recovers the constant on the highest power of a repeated factor, and the constant on any other distinct linear factor, directly. Every other constant needs coefficient matching afterward.

   Example

   2x+1=A(x-2)(x+1)+B(x+1)+C(x-2)^2. Cover-up at x=2: 5=3B\Rightarrow B=\tfrac53. Cover-up at x=-1: -1=9C\Rightarrow C=-\tfrac19. Matching the x^2 coefficient (which cover-up can't reach): 0=A+C\Rightarrow A=\tfrac19.

   • the factor is a repeated linear factor — cover-up gets the highest power's constant directly; match coefficients for the rest, as above.
   • the factor is an irreducible quadratic — there's no root to cover up at all; every constant in that term comes from matching coefficients.

   Why: covering up x=r on a repeated factor still makes every lower power's term blow up along with the one being solved for, so only the highest power is recoverable that way.

4. 4

   A repeated-factor term with power \geq2 integrates by the power rule, not the log rule. An irreducible-quadratic term splits into a log piece (from the derivative-matching part of the numerator) and an arctangent piece.

   Example

   \int\frac{2x+1}{(x-2)^2(x+1)}\,dx = \frac19\ln|x-2| - \frac{5}{3(x-2)} - \frac19\ln|x+1| + C

   Why: \dfrac{1}{(x-r)^2} antidifferentiates to -\dfrac{1}{x-r}, a power-rule form; treating it as a log term is the single most common error at this step.