Inverse Laplace Transform $\frac{p+2}{16((p+2)^2 + 4)}$ How do you work out the inverse laplace transform of
$$\dfrac{p+2}{16((p+2)^2 + 4)}$$
I know the $p+2$ is $e^{-2x}$ but what is the inverse of $(p+2)^2 +4$ ?
 A: First step: Assume you are given $$\frac{s}{s^2+4}$$ Second step: Solve $\mathcal{L}^{-1}\left(\frac{s}{s^2+4}\right)$ by this fact that $\mathcal{L}(\cos at)=\frac{s}{s^2+a^2}$. 
Third step: Note that multiplying the function $f(t)$ by $\exp(kt)$, make its L.T. to be $$\mathcal{F}(s)|_{s\to s-k}$$
A: Hint: Laplace transform of $e^c\cos(ax)$ is given by
$$  {\frac {s-c}{ \left( s-c \right) ^{2}+{a}^{2}}} $$
A: ok see here.....
you can write :  the numerator as :   (P - (-2))
and the denominator as:                ((P - (-2))^2) + (2)^2
Then you will get something llike this:      (p - (-2)) /  (((p - (-2))^2) + (2)^2)  -------this will be the total fraction!
and now  apply the fommula :      (e^(at))*(cos(bt))  =        (s-a) / (((s-a)^2) - (b^2))     ------- (note that for this problem you have P in place of S and you have to manipulate the signs and apply the formula thats all... )
where,    a= (-2) and  b= 2.    ------according to your problem.
And if we compare with the formula we will get the answer as:   (e^(-2t))*(Cos(2t))
Hope the discussion will help you....best of luck.
A: Use: $\color{red}{L^{-1}\{F(s-a)\}=e^{at}f(t)}$
$$L^{-1}\left\{\frac{1}{16}\cdot \frac{s}{s^2+2^2}\right\}=\frac{1}{16}\cos \left(2t\right)$$
So
$$L^{-1}\left\{\frac{s+2}{16\left(\left(s+2\right)^2+4\right)}\right\}=e^{-2t}\frac{1}{16}\cos \left(2t\right)$$
