14 Higher Engineering Mathematics
The denominator factorizes as (x − 1) (x + 3) and the
numerator is of less degree than the denominator. Thus
11 − 3x
x 2 + 2x − 3
may be resolved into partial fractions.
Let
11 − 3x
x 2 + 2x − 3
≡
11 − 3x
(x − 1)(x + 3)
≡
A
(x − 1)
+
B
(x + 3)
where A and B are constants to be determined,
i.e.
11 − 3x
(x − 1)(x + 3)
≡
A(x + 3) + B(x − 1)
(x − 1)(x + 3)
,
by algebraic addition.
Since the denominators are the same on each side
of the identity then the numerators are equal to each
other.
Thus, 11 −3x ≡ A(x + 3) + B(x − 1)
To determine constants A and B, values of x are chosen
to make the term in A or B equal to zero.
When x = 1, then
11 −3(1) ≡ A(1 + 3) + B(0)
i.e.
8 = 4 A
i.e.
A = 2
When x = −3, then
11 −3(−3) ≡ A(0) + B(−3 −1)
i.e.
20 = −4B
i.e.
B = −5
Thus
11 − 3x
x 2 + 2x − 3
≡
2
(x − 1)
+
−5
(x + 3)
≡
2
(x − 1)
−
5
(x + 3)
Check:
2
(x − 1)
−
5
(x + 3)
=
2(x + 3) − 5(x − 1)
(x − 1)(x + 3)
=
11 − 3x
x 2 + 2x − 3
Problem 2. Convert
2x 2 − 9x − 35
(x + 1)(x − 2)(x + 3)
into
the sum of three partial fractions.
Let
2x 2 − 9x − 35
(x + 1)(x − 2)(x + 3)
≡
A
(x + 1)
+
B
(x − 2)
+
C
(x + 3)
≡
A(x − 2)(x + 3) + B(x + 1)(x + 3)
+ C(x + 1)(x − 2)
(x + 1)(x − 2)(x + 3)
by algebraic addition.
Equating the numerators gives:
2x
2
− 9x − 35 ≡ A(x − 2)(x + 3)
+ B(x + 1)(x + 3) + C(x + 1)(x − 2)
Let x = − 1. Then
2(−1) 2 − 9(−1) − 35 ≡ A(−3)(2)
+ B(0)(2) +C(0)(−3)
i.e.
−24 =−6 A
i.e.
A =
−24
−6
= 4
Let x = 2. Then
2(2) 2 − 9(2) − 35 ≡ A(0)(5) + B(3)(5) + C(3)(0)
i.e.
−45 = 15B
i.e.
B =
−45
15
= −3
Let x = − 3. Then
2(−3) 2 − 9(−3) − 35 ≡ A(−5)(0) + B(−2)(0)
+ C(−2)(−5)
i.e.
10 = 10C
i.e.
C = 1
Thus
2x 2 − 9x − 35
(x + 1)(x − 2)(x + 3)
≡
4
(x + 1)
−
3
(x − 2)
+
1
(x + 3)
Problem 3. Resolve
x 2 + 1
x 2 − 3x + 2
into partial
fractions.