Question : how do you find inverse Laplace transform of 1/(s^2+2s-3)?

Solution :

Given \[f(s) =\frac{1}{(s^2+2s-3)}\]

It’s factorization is given as

\[ f(s)=\frac{1}{(s+3)(s-1)} \]

\[=> f(s)=\frac{1}{4}\left(\frac{1}{(s-1)}- \frac{1}{(s+3)}\right) \]

Now using inverse Laplace transform both sides

\[L^-1\left(f(s)\right) = \frac{1}{4}\left(L^-1\left(\frac{1}{s-1}\right)-L^-1\left(\frac{1}{s+3}\right)\right)\]

By inverse Laplace transform formulas

\[L^-1\left(\frac{1}{s+a}\right)=e^{-at}, L^-1\left(\frac{1}{s-a}\right)=e^{at}\]

\[=> f(t)=\frac{1}{4}\left(e^{t}-e^{-3t}\right)\]

is the required inverse Laplace transform of the  f(s)