Inverse Laplace Transform

I still need more clarification on this question..I added my work, someone please help, Finals are next week and this is the only thing im stuck on..Thank you so much..http://answers.yahoo.com/question/index;_ylt=Ag61LpV3BaEQEE0qZmeGbh7ty6IX;_ylv=3?qid=20120416125409AAywUtS

Partial fractions for 1/(s(s²−9)) should be
1/(s(s²−9)) = A/s + B/(s−3) + C/(s+3) or
1/(s(s²−9)) = A/s + (Bs+C)/(s²−9)

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Using first equation, we get:
1/(s(s²−9)) = A/s + B/(s−3) + C/(s+3) or
1 = A(s²−9) + B s(s+3) + C s(s−3)

s = 0 -----> 1 = −9A -----> A = −1/9
s = 3 -----> 1 = 18B -----> B = 1/18
s = −3 ---> 1 = 18C -----> C = 1/18

1/(s(s²−9)) = 1/18 (1/(s−3) + 1/(s+3) − 2/s)
. . . . . . . . . = 1/18 [((s+3)+(s−3))/(s²−9) − 2/s)]
. . . . . . . . . = 1/18 [2s/(s²−9) − 2/s)]
. . . . . . . . . = 1/9 (s/(s²−9) − 1/s))

which gives 1/9 (cosh(9t) − 1)

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Using second equation we get:

1/(s(s²−9)) = A/s + (Bs+C)/(s²−9)
1 = A(s²−9) + (Bs+C) s

s = 0 -------> 1 = −9A -----> A = −1/9
s = 3 -------> 1 = 9B + 3C
s = −3 -----> 1 = 9B − 3C

Adding last two equations, we get
18B = 2 -----> B = 1/9 -----> C = 0

1/(s(s²−9)) = (1/9 s + 0)/(s²−9) + (−1/9)/s
. . . . . . . . . = 1/9 (s/(s²−9) − 1/s))

which gives 1/9 (cosh(9t) − 1)