Decomposing a proper algebraic fraction into partial fractions, including denominators with distinct linear factors and a repeated linear factor, and using partial fractions in integration and binomial expansion.

What this dot point is asking

OCR wants you to split a proper algebraic fraction into a sum of simpler fractions: with distinct linear factors in the denominator, and with a repeated linear factor. You must first divide if the fraction is improper, then use either substitution of strategic values or comparison of coefficients to find the unknowns, and you must recognise that partial fractions make a rational expression easy to integrate or to expand as a binomial series.

The answer

The idea

A single rational expression with a factorised denominator can be written as a sum of fractions, one over each factor. This reverses the process of adding fractions over a common denominator, and the simpler pieces are far easier to integrate or expand.

Finding the unknowns

Multiply both sides by the full denominator to clear fractions, giving an identity. Then either substitute clever values of $x$ that make a factor zero (the quickest route for linear factors) or compare coefficients of like powers.

Repeated factors

A repeated factor $(x - a)^2$ needs both $\dfrac{A}{x - a}$ and $\dfrac{B}{(x - a)^2}$ in the decomposition. Substituting $x = a$ gives the $B$ term directly; the remaining constant comes from a second substitution or by comparing coefficients.

Improper fractions

If the numerator's degree is at least the denominator's, the fraction is improper: divide first to get a polynomial plus a proper fraction, then split the proper part.

Examples in context

Why it helps integration

$\displaystyle\int \dfrac{5x - 4}{(x + 1)(x - 2)}\,dx$ looks hard, but once split it is $\displaystyle\int\left(\dfrac{3}{x + 1} + \dfrac{2}{x - 2}\right)dx = 3\ln|x + 1| + 2\ln|x - 2| + c$. Each piece integrates to a logarithm.

Why it helps binomial expansion

To expand $\dfrac{1}{(1 - x)(1 + 2x)}$ as a series, split it first, then expand each $\dfrac{A}{1 \pm kx}$ with the binomial series $(1 + u)^{-1}$. Combining gives the series for the original expression, valid where every individual expansion converges.

Try this

Q1. Express $\dfrac{7}{(x - 2)(x + 5)}$ in partial fractions. [3 marks]

• Cue. $7 = A(x + 5) + B(x - 2)$; $x = 2$ gives $A = 1$, $x = -5$ gives $B = -1$, so $\dfrac{1}{x - 2} - \dfrac{1}{x + 5}$.

Q2. Integrate $\displaystyle\int\dfrac{1}{x - 2} + \dfrac{2}{x + 1}\,dx$. [2 marks]

• Cue. $\ln|x - 2| + 2\ln|x + 1| + c$.