Break into 2 elementary inverse transforms
L^-1{s/(s^2+16} = cos(4t)
For the second multiply and divide by 4:
L^-1 {3/(s^2+16} = 3/4 L^(-1){4/s^2+16} = 3/4 sin(4t)
It would help if your expression was posted less ambiguously.
By s+3/s^2+16 do you mean s + (3/(s^2 + 16)), s + (3/s^2) + 16 or (s + 3)(s^2 + 16)??
That is, by s+3/s^2+16 do you mean , or ? (As you can see, learning some basic latex would also help ..... see http://www.mathhelpforum.com/math-he...-tutorial.html)