Understanding Engineering Mathematics

and use the fact that this must be an identity to get relations betweenAandB:

A+B= 1

−( 2 A+B)= 1

the solution of which reproducesA=−2,B=3.
There is a short cut to all this, called thecover-up rule. Many teachers frown on such
‘tricks’ because they can be used blindly without real understanding. I certainly don’t
advocate that. Rather, they should be evidence of your mastery of the material, enabling
you to save time. In this case the trick is really a mental implementation of the first method
described above. We write

x+ 1

(x− 1 )(x− 2 )

≡

x− 1 x− 2

and fill in the numerators as follows. The numerator of the

1

x− 1

term is the value of

the left-hand side obtained by ‘covering up’, or ignoring, thex−1 factor and putting
x− 1 =0, i.e.x=1 in the remaining expression. This gives

x + 1

(x – 1) (x – 2)x = 1

= –2

as found forA.
Similarly, forBwe get

x + 1

(x – 1) (x – 2)x = 2

B = = 3

as before. With practice this may be done mentally and the result

x+ 1

(x− 1 )(x− 2 )

≡

− 2

x− 1

+

3

x− 2

written down immediately.
More complicated rational functions may be similarly decomposed in terms of partial
fractions, but the cover up rule may not apply, and we may have to resort to a combination
of methods. The general forms for such partial fractions are:

p(x)

(x−a)(x−b)(x−c)

≡

A

x−a

+

B

x−b

+

C

x−c

p(x)

(x−a)(x^2 +b)

≡

A

x−a

+

Bx+C

x^2 +b

p(x)

(x−a)(x−b)^3

≡

A

x−a

+

B

x−b

+

C

(x−b)^2

+

D

(x−b)^3

with obvious generalisations.p(x)is any polynomial of degree at least one less than the
denominator. Note that the number of constantsA,B,...to determine on the right-hand