1.11 Partial fractions (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

What are partial fractions?

Adding fractions combines them: $\frac{2}{x-1} + \frac{3}{x+2} = \frac{5x+1}{(x-1)(x+2)}$ . Partial fractions reverses this, splitting one fraction into simpler ones.

In IB AA HL you only need the case where:

the denominator is a product of two distinct linear factors, $(ax+b)(cx+d)$ , and
the degree of the numerator is less than the degree of the denominator (a proper fraction), so the numerator is linear or a constant.

Then $\frac{px+q}{(ax+b)(cx+d)} \equiv \frac{A}{ax+b} + \frac{B}{cx+d}.$

Key termspartial fractionsproper fractiondistinct linear factors

Exam tip

Always factorise the denominator first: $x^2 + x - 2 = (x-1)(x+2)$ .

Section 2

Method 1: substitution (cover-up)

Multiply both sides by the denominator to get an identity (true for every $x$ ): $5x + 1 \equiv A(x+2) + B(x-1).$ Substitute the value that makes each bracket zero:

$x = 1$ : $6 = 3A$ , so $A = 2$ .
$x = -2$ : $-9 = -3B$ , so $B = 3$ .

So $\frac{5x+1}{x^2+x-2} \equiv \frac{2}{x-1} + \frac{3}{x+2}$ .

Key termsidentity

Common mistake

Pairing $A$ with its own factor: it is $A(x+2)$ , not $A(x-1)$ , because the $(x-1)$ cancels when you multiply $\frac{A}{x-1}$ by $(x-1)(x+2)$ .

Section 3

Method 2: comparing coefficients

Expand the identity and match coefficients of each power of $x$ : $5x + 1 \equiv (A + B)x + (2A - B).$ So $A + B = 5$ and $2A - B = 1$ . Adding: $3A = 6$ , $A = 2$ , then $B = 3$ .

This method is slower here but is a useful check, and it is how you find constants when a convenient substitution is awkward.

Key termscomparing coefficients

Exam tip

Check your answer by substituting a simple value such as $x = 0$ into both sides: $\frac{1}{-2} = \frac{2}{-1} + \frac{3}{2}$ gives $-\frac12 = -\frac12$ , as required.

Section 4

Using partial fractions: sums that telescope

Partial fractions turn some series into telescoping sums, where most terms cancel. Since $\frac{2}{r(r+2)} = \frac1r - \frac{1}{r+2}$ , $\sum_{r=1}^{n}\frac{2}{r(r+2)} = \left(1 - \tfrac13\right) + \left(\tfrac12 - \tfrac14\right) + \left(\tfrac13 - \tfrac15\right) + \cdots + \left(\tfrac1n - \tfrac{1}{n+2}\right).$ Only the first two positive terms and last two negative terms survive: $\frac32 - \frac{1}{n+1} - \frac{1}{n+2}$ .

Key termstelescoping sum

Common mistake

Writing only the first term and the last term. Write out at least the first two and last two brackets to see exactly which terms survive.

Section 5

Using partial fractions: differentiating and integrating

Each partial fraction is easy to differentiate or integrate: $\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{k}{x+a}\right) = -\frac{k}{(x+a)^2}, \qquad \int \frac{k}{x+a}\,\mathrm{d}x = k\ln|x+a| + c.$ For example, if $f(x) = \frac{1}{x+1} + \frac{2}{x+3}$ then $f'(x) = -\frac{1}{(x+1)^2} - \frac{2}{(x+3)^2} < 0$ , so $f$ is decreasing for $x > -1$ .

For the drug model $C(t) = \frac{20t}{(t+1)(t+4)} = \frac{80}{3(t+4)} - \frac{20}{3(t+1)}$ , $\int_0^8 C(t)\,\mathrm{d}t = \frac{20}{3}\left[4\ln(t+4) - \ln(t+1)\right]_0^8 = \frac{40}{3}\ln3.$

Common mistake

Forgetting the coefficient of $x$ when integrating: $\int\frac{1}{2x+1}\,\mathrm{d}x = \frac12\ln|2x+1| + c$ .

Must know

Only two distinct linear factors, and a numerator of lower degree, are examined.
Write $\frac{px+q}{(x-a)(x-b)} \equiv \frac{A}{x-a} + \frac{B}{x-b}$ , then clear fractions to get an identity.
Find $A$ and $B$ by substituting $x = a$ and $x = b$ , or by comparing coefficients.
Check by substituting another value of $x$ .
Uses: telescoping sums, differentiation, integration into logarithms.

That's the notes covered.

Carry on to the next subtopic.