Question

inverse Laplace transform of s+1/s^2s+1
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Answer 1

Step-by-step explanation:

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Answer 2

Let,

f(s)= 1/(s²+s+1)

f(s)=1/(s+1/2)²+(√3/2)²

Then,

L^-1[f(s)]=L^-1[1/(s+1/2)²+(√3/2)²]

L^-1[f(s)]= e^-t/2.L^-1[1/[s²+(√3/2)²]]

L^-1[f(s)]=e^-t/2.2/√3.sin(√3t/2)

L^-1[1/s²+s+1]= 2/√3.e^-t/2.sin(√3t/2)