工數之inverse transforms

Q1:
L{cost}= s/(s^2 +1)
L{ exp(-t) cos(t)}= (s+1)/[(s+1)^2 + 1]= (s+1)/(s^2+2s+2)
L{ t exp(-t) cos(t) } = - d/ds [(s+1)/(s^2+2s+2) ]
= [ (2s+2)(s+1) - (s^2+2s+2) ] /(s^2+2s+2)^2
= (s^2+2s)/(s^2+2s+2)^2
Q2:
Property: L{ f(t) } = F(s) => L{ t f(t) } = - F'(s)
Let L{ f(t) } = ln[(s^2+1)/(s-1)^2 ]= ln(s^2+1) - 2 ln(s-1).
L{ t f(t) } = - 2s/(s^2+1) + 2/(s-1) = L{ -2cost + 2e^t }
=> f(t)= 2( e^t - cost )/ t