We recognize the question can be written as:

`(s+b)/(s(s^2+2bs+b^2+a^2))` `=(s+b)/(s((s+b)^2+a^2))`

Once again, we will use Property (3). First, we have:

`Lap^{:-1:}{(s+b)/((s+b)^2+a^2)}` `=e^(-bt)cos at`

`Lap^{:-1:}{(s+b)/(s((s+b)^2+a^2))}` `=int_0^te^(-bt)cos at\ dt`

From our table of integrals, we have:

`inte^(au)cos bu\ du` `=(e^(au)(a\ cos bu+b\ sin bu))/(a^2+b^2)`

So our Laplace Inverse is given by:

`g(t) = int_0^te^(-bt)cos at\ dt`

`=[(e^(-bt)(-b\ cos at+a\ sin at))/(a^2+b^2)]_0^t`

`=(e^(-bt)(-b\ cos at+a\ sin at))/(a^2+b^2)-` `(-b/(a^2+b^2))`

`=(e^(-bt)(-b\ cos at+a\ sin at)+b)/(a^2+b^2)`

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