Find the inverse transform of the following function Please

Find the inverse transform of the following function (Please show work)
20) (2)/(s²+2s+10)

22) (e^-2s)/(s2)

23) (e^(pi)(s) ) /(s²+1)

Solution

20.)  First, we can complete the square on the denominator to get s^2 + 2s + 10 = (s + 1)^2 + 3^2,

f(t) = ILT{F(s)}

f(t) = ILT (2/[(s + 1)^2 + 3^2]}

f(t) = 2/3  e^(-t) sin(3t).

22.) By partial fractions,
1/(s^2 = A/s + B/s^2 .

Clearing denominators yields
1 = As + B

s = 0 ==> B = 1
s = 1 , A+B=1

B=1, A=0

So, L^(-1) {e^(-2s)/(s^2}
= L^(-1) {e^(-2s) [1/s^2]}
= (t-2)u(t-2)

23.) if F(s) = 1/(s^2+1)

thne f(t) = sin t

Recall L1[easF(s)](t)=L1[F(s)](ta)u(ta)=f(ta)u(ta)]

Apply that theorem to your F with a=- to obtain

sin(t+pi)(t+pi)