Understanding Engineering Mathematics

2.2.10 Partial fractions

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39 77 ➤

Often we have to reverse the process of taking the common denominator, i.e. split an
expression of the form

px+q

(x−a)(x−b)

(p,qnumbers)

into itspartial fractions

A

x−a

+

B

x−b

The expression is easier to differentiate and integrate in this form, for example. The method
used is best illustrated by an example. We effectively have to reverse the process of taking
the common denominator. But this is an example of a mathematical process that is much
easier to do than it is to undo!
Consider

x+ 1

(x− 1 )(x− 2 )

≡

A

x− 1

+

B

x− 2

where the object is to determineAandBon the right-hand side. There are a number of
ways of doing this, but they all implicitly depend on combining the right-hand side over
a common denominator and insisting that the numerators be identical on both sides:

A

x− 1

+

B

x− 2

≡

A(x− 2 )+B(x− 1 )

(x− 1 )(x− 2 )

≡

x+ 1

(x− 1 )(x− 2 )

This implies

A(x− 2 )+B(x− 1 )≡x+ 1 ( 1 )

There are now two ways to go, both relying on the fact that the result is anidentity(50

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)
and must therefore be true for all values ofx. In particular, it is true for

x= 1 , which givesA=− 2

and x= 2 , which givesB= 3

So

x+ 1

(x− 1 )(x− 2 )

≡

3

x− 2

−

2

x− 1

NB. Checking such results by recombining the right-hand side to confirm that you do
indeed get the left-hand side will help you to develop your algebraic skills.
Another means of findingAandBis to rewrite the identity (1) as

(A+B)x−( 2 A+B)≡x+ 1