# “Teach A Level Maths” Vol. 2: A2 Core Modules

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“Teach A Level Maths” Vol. 2: A2 Core Modules
46: Applications of Partial Fractions © Christine Crisp

There are 2 topics where partial fractions are useful.
finding some binomial expansions differentiating some algebraic fractions integrating some algebraic fractions

Algebraic fractions can be awkward to differentiate and changing them into partial fractions makes them much easier. e.g. Differentiate Solution: The partial fractions are We need the chain rule here.

Uses in Integration Problems
Expressions Which Give Ln`s If the Top is Gradient of Bottom Ans = ln(bottom)

Top is NOT Gradient of Bottom so FIX it.
The gradient of the bottom is 4 so put a 4 on the top line and divide by 2 so that the question has not changed The gradient of the bottom is 15x2 so put a 15x2 on the top line and  by 4/15 so that the question has not changed

If the Top is Gradient of Bottom Ans = ln(bottom) This constant . . .

This constant . . . is usually replaced by , so we get which can be tidied up using the 1st law of logs to give

We’ll take (b) as our 1st example using partial fractions.
e.g.1 Find Factorise the bottom and express as 2 partial fractions Solution: so,

It’s safer to separate the terms and adjust the constants for each. (-2)

The 3 logs can now be simplified:
using the 3rd law: using the 1st and 2nd laws:

e.g. 2 Write as the sum of 3 partial fractions and then find where
Solution:

The partial fractions are:
so, The 2nd and 3rd integrals give logs but the 1st doesn’t. The 1st is of the form (This integral comes up often so it’s worth remembering)

so, and Since one term doesn’t give a log, we don’t gain by using instead of C.

e.g.3 Express as a single log
Solution: Multiply by : So:

It makes the integration easier if we now write

SUMMARY Rational functions can be integrated if they are of the form or if they can be written as partial fractions with a linear denominator the fraction will integrate to a log with a repeated factor the fraction will integrate to another fraction e.g. If all terms integrate to logs, replaces C

Exercises 1. Express the following as a single logarithm (a) 2. Find (b)

1(a) Solutions:

(b) The partial fractions are so,

2.

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

There are 2 topics where partial fractions are useful.
finding some binomial expansions integrating some algebraic fractions differentiating some algebraic fractions

SUMMARY To find a binomial expansion for an algebraic fraction with a denominator that factorises into linear factors: Find the partial fractions. For each partial fraction, write the denominator with a negative power. Expand each part and write the values for which the series converges. Find the validity of the entire expansion by choosing the most stringent of the restrictions on x. Combine the expansions. For expressions of the form , remove from the brackets.

We expand the 2 fractions separately. We have:
Replace n by e.g.

The 2nd fraction is For the binomial we must have , so we take outside the brackets: so we can save time by replacing x by in that. The 1st fraction gave so,

So,

Algebraic fractions can be awkward to differentiate and changing them into partial fractions makes them much easier. e.g. Differentiate Solution: The partial fractions are We need the chain rule here.

Rational functions can be integrated if
SUMMARY they are of the form or if they can be written as partial fractions fractions with a linear denominator will integrate to logs fractions with a repeated factor will integrate to another fraction e.g. If all terms integrate to logs, replaces C

and There’s an important difference. (a) is of the form and can be integrated directly. (b) is not of this form and cannot be converted to it by using a multiplying constant. Instead, we use partial fractions. Consider

There is one thing it’s useful to notice about (a) before we move to (b)
This constant . . . is usually replaced by , so we get which can be tidied up using the 1st law of logs to give

e.g.1 Find Solution: We found earlier that so, Both terms give us log integrals. It’s safer to separate the terms and adjust the constants for each.

(-2) The 3 logs can now be simplified: using the 3rd law:
using the 1st and 2nd laws:

e.g.2 Find The 2nd and 3rd integrals give logs but the 1st doesn’t. The 1st is of the form Solution: (This integral comes up often so it’s worth remembering)

so, and Since one term doesn’t give a log, we don’t gain by using instead of C.