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Partial Fractions
.

then by means of Theorem 2 the partial fraction expansion in this case is

2
s − s + 9 A Bs + C
≡ +
3
2
s + 9s s s + 9
where A, B and C need to be determined. We do this by first clearing
denominators
2
2
s − s + 9 ≡ A(s + 9) + (Bs + C)s.
Comparing coefficients of like powers – rearranging the right side of the
last equality gives

2
2
s − s + 9 ≡ (A + B)s + Cs + 9A,
or
 
 A + B = 1  A = 1
C = −1 ⇒ B = 0
 
9A = 9 C = −1
Hence,

[ 2 ] [ ] [ 1 ] [ ] 1 [ 3 ]
1
1
s − s + 9
−1 −1 −1 −1 −1
L = L − L = L − L =
2
2
3
s + 9s s s + 9 s 3 s + 9
1
= 1 − sin 3t,
3
. . . . which is our answer.
.
[ s − 8 ]
−1
Example 2.33, Find L .
2
s + 4s + 29
. . . . .
.
Solution. First complete the square of the denominator

2
2
2
s + 4s + 13 = s + 4s + 4 + 25 = (s + 2) + 25.
We need to determine

[ s − 8 ] [ (s + 2) − 10 ]
−1 −1
L = L =
2
2
s + 4s + 29 (s + 2) + 5 2
[ s + 2 ] [ 5 ]
−1 −1 −2t
= L − 2 · L = e cos 5t − 2 sin 5t.
2
2
(s + 2) + 5 2 (s + 2) + 5 2
. . . .
In many cases, F(s) is the quotient of two polynomials with real coefficients.
If the numerator polynomials is of the same or higher degree than the
denominator polynomial, first divide the numerator polynomial by the

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