Inverse Laplace transform

Question

L^-1(s/(s+1))

Answer 1

Laplace Transform is defined as  ℒ{f(t)}=∫₀^∞e^−stf(t) =G(s) dt and its inverse 

ℒ^-1{(s/(s+1))}=(1/2πi)∫_-i∞^i∞e^sts/(s+1)ds

=(1/2πi)∫_-i∞i∞e^st[1-1/(s+1)]ds=

=(1/2πi)∫_-i∞^i∞e^stds-(1/2πi)∫_-i∞^i∞e^st/(s+1)ds=-e^t+δ(t)