Integration by partial fractions (WACE Year 12 Mathematics Specialist Unit 4)

What this dot point is asking

SCSA wants you to decompose a rational integrand into partial fractions and integrate, including reducing an improper fraction by division first.

When and how to decompose

Partial fractions apply to a proper rational function (numerator degree less than denominator degree) whose denominator factors. For distinct linear factors, set up one constant over each factor:

Multiply through by the denominator and solve for the constants, either by substituting the roots $x = a$ and $x = b$ (the cover-up method) or by equating coefficients.

The cover-up method in detail

The cover-up method is the fastest way to find the constants over distinct linear factors. To find $A$ in $\dfrac{px + q}{(x - a)(x - b)}$, mentally cover the $(x - a)$ factor in the denominator and evaluate the remaining expression at $x = a$, giving $A = \dfrac{pa + q}{a - b}$. The justification is that multiplying both sides by $(x - a)$ and then setting $x = a$ kills every term except the one over $(x - a)$. It works whenever the factor is linear and appears to the first power, which covers the standard SCSA cases. For repeated or quadratic factors the cover-up trick only finds some of the constants, so equating coefficients is needed to complete the system.

Repeated and irreducible quadratic factors

Although distinct linear factors are the most common case, SCSA also expects you to recognise two harder forms. A repeated linear factor $(x - a)^2$ needs two terms, $\dfrac{A}{x - a} + \dfrac{B}{(x - a)^2}$, because a single term cannot capture both behaviours. An irreducible quadratic factor $x^2 + c$ in the denominator needs a linear numerator, $\dfrac{Bx + C}{x^2 + c}$, which integrates to a combination of a logarithm (from the $Bx$ part) and an inverse tangent (from the $C$ part). Setting up the correct template before solving for constants is half the battle.

Improper fractions: divide first

Integrating the pieces

Each term $\tfrac{A}{x - a}$ integrates to $A\ln|x - a|$. So the integral of a decomposed rational function is a sum of logarithms (plus any polynomial part from division), combined into a single logarithm where convenient using log laws.

Why the technique works and where it leads

Partial fractions matter because the antiderivative of $\dfrac{1}{x - a}$ is a logarithm, a function the other techniques cannot easily produce from a rational integrand. Splitting a complicated fraction into a sum of single-factor pieces converts an integral with no obvious antiderivative into a sum of standard logarithmic integrals. This is precisely the step that makes the logistic differential equation solvable: separating variables leaves $\displaystyle\int \dfrac{dP}{P(1 - P/M)}$, whose integrand is decomposed by partial fractions into $\dfrac{1}{P}$ and a term in $\dfrac{1}{1 - P/M}$, each of which integrates to a logarithm. Mastery of partial fractions therefore underpins a substantial part of the differential-equations work later in Unit 4.