Recipe

Integrate a rational function with distinct linear factors

Use when: the integrand is a proper rational function whose denominator factors into distinct (non-repeated) linear factors

Goal: the rational function rewritten as a sum of simple A/(x-r) terms, each integrating directly to a logarithm

Taught in Partial fractions — distinct linear factors · a later section extends this recipe with more cases

1. 1

   Check that the numerator's degree is less than the denominator's; if not, do polynomial long division first. Factor the denominator completely into distinct linear factors.

   Example

   \displaystyle\int\frac{3x+5}{x^2-1}\,dx: the numerator has degree 1, the denominator degree 2, so the fraction is already proper. Factor: x^2-1=(x-1)(x+1).

   Why: partial fractions only decomposes a proper fraction; an improper one has to be reduced first or the decomposition doesn't add up correctly.

2. 2

   Write the fraction as a sum with one unknown constant over each linear factor.

   Example

   \frac{3x+5}{(x-1)(x+1)} = \frac{A}{x-1}+\frac{B}{x+1}

   Why: this is the shape the decomposition is guaranteed to take whenever every factor is linear and distinct.

3. 3

   Multiply both sides by the full denominator, then plug in each root in turn to isolate one constant at a time.

   Example

   3x+5=A(x+1)+B(x-1). At x=1: 8=2A\Rightarrow A=4. At x=-1: 2=-2B\Rightarrow B=-1.

   Why: plugging in x=r_i makes every other term vanish, since it's multiplied by (r_i-r_i)=0, leaving exactly one constant to solve for.

4. 4

   Each term A/(x-r) integrates directly to A\ln|x-r|.

   Example

   \int\frac{3x+5}{x^2-1}\,dx = 4\ln|x-1| - \ln|x+1| + C

   Why: once split, each term is the simplest possible reverse power rule, in logarithmic form, so no further technique is needed.

5. 5

   Substitute a convenient x into both the original fraction and the decomposition to confirm the constants are correct.

   Example

   At x=0: original is \dfrac{5}{-1}=-5; decomposition is \dfrac{4}{-1}+\dfrac{-1}{1}=-4-1=-5. They match.

   Why: this catches a sign error in cover-up before it propagates into the final logarithms.