Integration by partial fractions in HSC Mathematics Extension 2: decomposing rational functions over linear, repeated and irreducible quadratic factors and integrating

What this dot point is asking

NESA wants you to integrate rational functions (a polynomial divided by a polynomial) by first splitting them into simpler partial fractions. You must recognise the three cases (distinct linear factors, repeated linear factors, and irreducible quadratic factors), write the correct decomposition form for each, find the unknown constants, and integrate the resulting pieces into logarithms and arctangents. The skill is examined both as a standalone "express as partial fractions" question and as the engine inside a larger integral.

When to use partial fractions

The method applies to proper rational functions, where the numerator has lower degree than the denominator. That degree condition is not a technicality, it is what guarantees a decomposition exists with constant (and linear) numerators. If the function is improper (numerator degree greater than or equal to denominator degree), you must first divide using polynomial long division to extract a polynomial part plus a proper remainder, then decompose only the remainder. Skipping the division on an improper fraction produces a system of equations with no solution, which is a sure sign you have started in the wrong place.

The deeper reason the method is worth the effort: a single complicated rational function has no obvious antiderivative, but each partial fraction is a textbook standard form. So decomposition converts a hard problem into a sum of easy ones, and you can predict the shape of the answer (logarithms from linear factors, arctangents from irreducible quadratics) before doing any integrating.

Distinct linear factors

If the denominator factorises into distinct linear factors, write one fraction per factor:

Multiply through by $(x - a)(x - b)$ to clear denominators, then either equate coefficients or substitute convenient values of $x$. Each term integrates as $\int \dfrac{A}{x - a}\,dx = A\ln|x - a| + C$.

Repeated linear factors

A repeated factor $(x - a)^2$ needs a separate term for every power up to the multiplicity, not just one term:

The reason both terms are needed is a counting argument: a proper fraction over $(x - a)^2$ has two degrees of freedom in its numerator, so the decomposition must also carry two unknowns. Using only $\dfrac{B}{(x-a)^2}$ throws one away and the system fails. A cube $(x - a)^3$ would need three terms, up to $\dfrac{C}{(x - a)^3}$. The first term integrates to a logarithm; the higher terms integrate by the power rule, for example $\int \dfrac{B}{(x - a)^2}\,dx = -\dfrac{B}{x - a}$.

Irreducible quadratic factors

A quadratic $x^2 + px + q$ with no real roots (negative discriminant, $p^2 - 4q < 0$) cannot be split into real linear factors, so it gets a full linear numerator:

A constant numerator is not enough here, for the same degrees-of-freedom reason: a proper fraction over a quadratic has a numerator of degree up to one, so you need both $B$ and $C$. Such a term integrates in two pieces: split it into a part proportional to the derivative $2x + p$ of the denominator (which integrates to a logarithm) plus a leftover constant, which after completing the square integrates to an arctangent.

The cover-up method

For distinct linear factors there is a shortcut that avoids solving a system. To find the constant over $(x - a)$, multiply the whole identity by $(x - a)$ and then set $x = a$: every other term still has a factor of $(x - a)$ in it and vanishes, leaving the constant alone. In practice you "cover up" the factor $(x - a)$ in the original denominator and evaluate what remains at $x = a$. For $\dfrac{5x - 3}{(x + 1)(x - 3)}$, the constant over $(x + 1)$ is found by covering $(x + 1)$ and putting $x = -1$ into $\dfrac{5x - 3}{x - 3} = \dfrac{-8}{-4} = 2$. The cover-up method is fast and reliable for simple linear factors, but it only isolates constants over single linear factors; for repeated or irreducible-quadratic blocks you still need to equate coefficients for the remaining unknowns.

Finding the unknown constants

Two methods recover the constants $A$, $B$, $C$ after you clear the denominators, and good answers mix them. The cover-up (substitution) method substitutes the value of $x$ that makes one factor zero, instantly isolating the constant over that factor; it is fastest for distinct linear factors. The equate-coefficients method expands the right side and matches the coefficient of each power of $x$ on both sides, producing a small linear system; it is the reliable route when irreducible quadratics or repeated factors are involved, since not every constant can be isolated by a single substitution. The efficient workflow is to cover up the linear factors first, then equate one or two coefficients (often the highest power and the constant term) to pin down the rest. Always verify with a check value of $x$, as in the worked examples below, before integrating.

Integrating each piece

Once decomposed, each fragment maps to a standard integral, and recognising the target shape in advance is a useful sanity check on your decomposition.

• $\displaystyle\int \frac{A}{x - a}\,dx = A\ln|x - a| + C$ (a logarithm; keep the absolute value).
• $\displaystyle\int \frac{B}{(x - a)^2}\,dx = -\frac{B}{x - a} + C$ (power rule, no logarithm).
• $\displaystyle\int \frac{Bx + C}{x^2 + px + q}\,dx$: write the numerator as $\tfrac{B}{2}(2x + p)$ plus a constant. The first part gives $\tfrac{B}{2}\ln(x^2 + px + q)$ (no absolute value, since an irreducible quadratic is always positive); the leftover constant over the completed square gives an arctangent.

So the final answer to a partial-fractions integral is always a combination of logarithms and arctangents (plus possibly a polynomial, if the original fraction was improper).

How exam questions ask about partial fractions

• "Express $\dfrac{\ldots}{\ldots}$ as a sum of partial fractions." Decomposition only, no integration. Choose the correct form for each factor and find all constants; finish with the constants substituted in.
• "Use partial fractions to find $\int \ldots$." Decompose first, then integrate term by term to logs and arctangents, with $+C$.
• "Hence find $\int_a^b \ldots$" (definite). Decompose, integrate, then substitute the limits; the modulus inside the logs keeps everything real over the interval.
• A denominator with an irreducible quadratic ($x^2 + 1$, $x^2 + x + 1$). Use a linear numerator $Bx + C$; expect an arctangent in the answer.
• A repeated factor $(x - a)^2$ or $(x + a)^2$. Include both $\dfrac{A}{x - a}$ and $\dfrac{B}{(x - a)^2}$; the second integrates without a log.
• An improper-looking integrand (numerator degree $\ge$ denominator degree). Do polynomial long division first, then decompose the remainder.