18 Higher Engineering Mathematics
The denominator is a combination of a quadratic factor,
(x 2 + 2), which does not factorize without introducing imaginary surd terms, and a linear factor, (x + 1).
Let,
7x 2 + 5x + 13
(x 2 + 2)(x + 1)
≡
Ax + B
(x 2 + 2)
+
C
(x + 1)
≡
(Ax + B)(x + 1) + C(x 2 + 2)
(x 2 + 2)(x + 1)
Equating numerators gives:
7x
2
+ 5x + 13 ≡ (Ax + B)(x + 1) + C(x
2
+ 2) (1)
Let x =−1. Then
7(−1) 2 + 5(−1) + 13 ≡ (Ax + B)(0) + C(1 + 2)
i.e.
15 = 3C
i.e.
C = 5
Identity (1) may be expanded as:
7x
2
+ 5x + 13 ≡ Ax
2
+ Ax + Bx + B + Cx
2
+ 2C
Equating the coefficients of x 2 terms gives:
7 = A + C, and since C = 5, A = 2
Equating the coefficients of x terms gives:
5 = A + B, and since A = 2, B = 3
[Check: equating the constant terms gives:
13 = B + 2C
When B = 3 and C = 5,
B + 2C = 3 + 10 = 13 = LHS]
Hence
7x 2 + 5x + 13
(x 2 + 2)(x + 1)
≡
2x + 3
( x 2 + 2)
+
5
(x + 1)
Problem 9. Resolve
3 + 6x + 4x 2 − 2x 3
x 2 (x 2 + 3)
into
partial fractions.
Terms such as x 2 may be treated as (x + 0) 2 , i.e. they
are repeated linear factors.
Let
3 + 6x + 4x 2 − 2x 3
x 2 (x 2 + 3)
≡
A
x
+
B
x 2 +
Cx + D
(x 2 + 3)
≡
Ax(x 2 + 3) + B(x 2 + 3) + (Cx + D)x 2
x 2 (x 2 + 3)
Equating the numerators gives:
3 + 6x + 4x
2
− 2x
3
≡ Ax(x
2
+ 3) + B(x
2
+ 3)
+ (Cx + D)x
2
≡ Ax
3
+ 3 Ax + Bx
2
+ 3B
+ Cx
3
+ Dx
2
Let x = 0. Then 3 = 3B
i.e.
B = 1
Equating the coefficients of x 3 terms gives:
−2 = A + C
(1)
Equating the coefficients of x
2 terms gives:
4 = B + D
Since B = 1, D = 3
Equating the coefficients of x terms gives:
6 = 3 A
i.e.
A = 2
From equation (1), since A = 2, C = −4
Hence
3 + 6 x + 4x 2 − 2 x 3
x 2 (x 2 + 3)
≡
2
x
+
1
x 2 +
−4x + 3
x 2 + 3
≡
2
x
+
1
x 2 +
3 − 4x
x 2 + 3
Now try the following exercise
Exercise 10 Further problems on partial
fractions with quadratic factors
1.
x
2
− x − 13
(x 2 + 7)(x − 2)
2x + 3
(x 2 + 7)
−
1
(x − 2)
2.
6x − 5
(x − 4)(x 2 + 3)
1
(x − 4)
+
2 − x
(x 2 + 3)
3.
15 + 5x + 5x 2 − 4x 3
x 2 (x 2 + 5)
1
x
+
3
x 2 +
2 − 5x
(x 2 + 5)