Finding laplace inverse of $\frac{1}{(p+2)^2(p^2+\omega^2)}$ I am finding the laplace inverse of the below :-
$$\frac{1}{(p+2)^2(p^2+\omega^2)}$$
I can apply convolution theorem by taking $$f(p)=\frac{1}{(p+2)^2}, g(p)=\frac{1} 
{(p^2+\omega^2)}$$
and calculating the function which yield the above functions on laplace transform  i.e. essentially ,
$$F(t)=e^{-2t}t,G(t)=\frac{\sin{\omega}t}{\omega}$$
then solve by routine integration .
But this seems a little cumbersome . Is there a better way to solve the above ?
Thanks for reading !
 A: Hint:
Use  Partial Fraction Decomposition
$$\dfrac1{(p^2+w^2)(p+2)^2}=\dfrac a{p+2}+\dfrac b{(p+2)^2}+\dfrac{cp+d}{p^2+w^2}$$
Multiply both sides by $(p^2+w^2)(p+2)^2$
and  use the coefficients of different exponents of $p$ to find $a,b,c,d$
Finally use the  standard formulae
A: The inverse Laplace transformation is define as $f(t)=\frac{1}{2\pi{i}}\int_{a-i\infty}^{a+i\infty}F(p)e^{pt}dp$, where integral is taken in the complex plane along some line $(a-i\infty,a+i\infty)$, where $a$ is some real number (greater than the real part of all singularities of $F(p)$)
https://en.wikipedia.org/wiki/Inverse_Laplace_transform
We can close the contour, integrating along a part of the circle of big radius $R$.
Integral along this circle $\to0$ at $R\to\infty$  if $|F(p)|\to 0$ at $|p|\to\infty$ ; this is the case for $F(p)=\frac{1}{(p+2)^2(p^2+\omega^2)}$

$$f(t)=\oint_C=\frac{2\pi{i}}{2\pi{i}}\sum_{\text{all poles}}Res\frac{e^{pt}}{(p+2)^2(p^2+\omega^2)}=\frac{e^{i\omega t}}{2i\omega(2+i\omega)^2}+\frac{e^{-i\omega t}}{-2i\omega(2-i\omega)^2}+\frac{d}{dp}\bigl(\frac{e^{pt}}{p^2+\omega^2}\bigr)|_{p=-2}$$
$$=\frac{1}{\omega}\Im\frac{e^{i\omega t}}{(2+i\omega)^2}+\frac{e^{-2t}}{4+\omega^2}\bigl(t+\frac{4}{4+\omega^2}\bigr)$$
$$=\frac{(4-\omega^2)\sin \omega t-4\omega\cos\omega t}{\omega(4+\omega^2)^2}+\frac{e^{-2t}}{4+\omega^2}\bigl(t+\frac{4}{4+\omega^2}\bigr)$$
Please, check.
