To find Inverse Laplace of $\,F(s)=\log\dfrac{s+1}{(s+2)(s+3)}$

Question

To find Inverse Laplace of $$F(s)= \log\frac{s+1}{(s+2)(s+3)}.$$

I have tried to use shifting theorems, but of no use. Should I apply series for log and take inverse laplace of individual terms, if yes under which conditions series is valid. I their any other way to do this?

Thanks

Answer 1

First of all, $\log\dfrac{ab}c=\log a+\log b-\log c$

$$L^{-1}\log(s-a)=-\dfrac1t L^{-1}\left(\dfrac{d\{\log(s-a)\}}{ds}\right)=-\dfrac1tL^{-1}\left(\dfrac1{s-a}\right)$$

$$L^{-1}\left(\dfrac1{s-a}\right)=e^{at}$$

Answer 2

F(s)=log[(s+1)/(s+2)(s+3)]

f(s)=log(s+1)-log(s+2)-log(s+3)

differentiating with respect to s

f'(s)= (1/s+1)-(1/s+2)-(1/s+3)

taking inverse laplace

-t*f(t)= (e^-1)-(e^-2)-(e^-3)

f(t)= -[(e^-1)-(e^-2)-(e^-3)]/t