A function f (t) that has the given Laplace transform F (s).

Given:
Laplace transform F (s) of a function f (t) defined on an interval $[0,\mathrm{\infty })-F(s)=\frac{2}{s(s^2+16)}$
Calculation:
Given,
$F(s)=\frac{2}{s(s^2+16)}$
Partial fraction decomposition,
$F(s)=\frac{2}{s(s^2+16)}$
Approach:
The
$=\frac{A}{s}+\frac{Bs+C}{s^2+16}$
$2=A(s^2+16)+s(Bs+C)$
$=s^2(A+B)+Cs+16A$
Compare the Left hand side and Right hand side,
$A=\frac{1}{8}$
$B=-\frac{1}{8}$
$C=0$
Substitute -
$F(s)=\frac{1}{8s}-\frac{s}{8(s^2+16)}$
Laplace transform,
$L[f]={\int }_0^{\mathrm{\infty }}{e}^{-s}tf(t)dt$
$=F(s)$
$=\frac{2}{s(s^2+16)}$
$=\frac{1}{8s}-\frac{s}{8(s^2+16)}$
Inverse Laplace transform-
$L^{-1}[F](t)=L^{-1}[\frac{2}{s(s^2+16)}]$
$=L^{-1}[\frac{1}{8s}-\frac{s}{8(s^2+16)}]$
$=\frac{1}{8}{L}^{-1}\left[\frac{1}{s}\right]-\frac{1}{8}{L}^{-1}\left[\frac{s}{s^2+4^2}\right]$
$=\frac{1}{8}-\frac{1}{8}\cos \left(4t\right)$
$L^{-1}\left[F\right]\left(t\right)=\frac{1}{8}(1-\cos \left(4t\right))$
$=f(t)$
Conclusion:
Hence, a function f (t) that has the given Laplace transform
$F\left(s\right)=\frac{2}{s(s^2+16)}is\frac{1}{8}(1-\cos \left(4t\right)).$