Example 1. Find the original $f(t)$, if

$$
F(p)=\frac{p+2}{(p+1)(p-2)\left(p^{2}+4\right)}
$$
Solution. First method. Represent $F(p)$ as a sum of simple fractions

$$
\frac{p+2}{(p+1)(p-2)\left(p^{2}+4\right)}=\frac{A}{p+1}+\frac{B}{p-2}+\frac{C_{p}+D}{p^{2}+4}
$$

and find the undetermined coefficients $A, B, C, D$. We have

$$
p+2=A(p-2)\left(p^{2}+4\right)+B(p+1)\left(p^{2}+4\right)+\left(C_{p}+D\right)(p+1)(p-2)
$$

By setting $p=-1, p=2, p=2i$ in the last equation, we get $-15 A=1, 24 B=4$

$$
(2 C i+D)(2 i+1)(2 i-2)=2+2 i
$$

from which $A=-\frac{1}{15}, B=\frac{1}{6}, C=-\frac{1}{10}, D=-\frac{2}{5}$; thus,

$$
F(p)=-\frac{1}{15} \frac{1}{p+1}+\frac{1}{6} \frac{1}{p-2}-\frac{1}{10} \frac{p+4}{p^{2}+4}
$$

Finding the originals for each of the simple fractions and using the linearity property, we get

$$
f(t)=-\frac{1}{15} e^{-t}+\frac{1}{6} e^{2 t}-\frac{1}{10} \cos 2 t-\frac{1}{5} \sin 2 t
$$

Second method. Find the poles $p_{k}$ of the function $F(p)$. They coincide with the zeros of the denominator $B(p)=(p+1)(p-2)\left(p^{2}+4\right)$. Thus, the image $F(p)$ has four simple poles $p_{1}=-1, p_{2}=2, p_{3}=2i, p_{4}=-2i$. Using formula (17), we get the original

$$
\begin{aligned}
f(t) & =\sum_{k=1}^{4} \frac{A\left(p_{k}\right)}{B^{\prime}\left(p_{k}\right)} e^{p_{k} t}=\sum_{k=1}^{4} \frac{p_{k}+2}{4 p_{k}^{3}-3 p_{k}^{2}+4 p-4} e^{p_{k} t}= \\
& =-\frac{1}{15} e^{-t}+\frac{1}{6} e^{2 t}+\frac{-1+2 i}{20} e^{2 i t}+\frac{-1-2 i}{20} e^{-2 i t}= \\
& =\frac{1}{6} e^{2 t}-\frac{1}{15} e^{-t}-\frac{1}{10} \cos 2 t-\frac{1}{5} \sin 2 t
\end{aligned}
$$