3.40. Find the originals of the following functions:
a) $\bar{f}(p)=\frac{p}{p^{2}-2 p+5}$
b) $\bar{f}(p)=\frac{1}{p^{3}-8} ;$
c) $\bar{f}(p)=\frac{p+1}{p(p-1)(p-2)(p-3)}$.
Solution. a) We use elementary techniques to decompose this fraction into the sum of fractions whose originals are known from the table:

$$
\frac{p}{p^{2}-2 p+5}=\frac{p-1+1}{(p-1)^{2}+4}=\frac{p-1}{(p-1)^{2}+4}+\frac{1}{(p-1)^{2}+4}
$$

(completing the square in the denominator).

By formulas 13,12 of Table 3.2, we have:

$$
\begin{aligned}
& \frac{p-1}{(p-1)^{2}+4} \doteqdot e^{t} \cos 2 t \quad(\hat{\lambda}=1,(\omega=2) \\
& \frac{1}{(p-1)^{2}+4}=\frac{1}{2} \frac{2}{(p-1)^{2}+4} \doteqdot \frac{1}{2} e^{t} \sin 2 t \quad(\lambda=1, \omega=2)
\end{aligned}
$$

Therefore,

$$
\frac{p}{p^{2}-2 p+5} \doteqdot e^{t}\left(\cos 2 t+\frac{1}{2} \sin 2 t\right)
$$

b) In this example, we use elementary techniques for decomposition known from integral calculus. We decompose the given fraction into simpler fractions:

$$
\frac{1}{p^{3}+8}=\frac{A}{p-2}+\frac{B p+C}{p^{2}+2 p+4}
$$

To determine the coefficients, we have the identity:

$$
\begin{gathered}
1 \equiv A\left(p^{2}+2 p+4\right)+(B p+C)(p-2) \\
1 \equiv(A+B) p^{2}+(2 A-2 B+C) p+(4 A-2 C) .
\end{gathered}
$$

This identity is clearly satisfied for those triples of numbers $(A, B, C)$ which are solutions to the system:

$$
\left\{\begin{array}{l}
A+B=0 \\
2 A-2 B+C=0 \\
4 A-2 C=1
\end{array}\right.
$$

Solving this, we find that:

$$
(A, B, C)=\left(\frac{1}{12} ;-\frac{1}{12} ;-\frac{1}{3}\right) .
$$

Therefore:

$$
\begin{gathered}
\frac{1}{p^{3}+8}=\frac{1}{12}\left(\frac{1}{p-2}-\frac{p+4}{p^{2}+2 p+4}\right)= \\
=\frac{1}{12}\left(\frac{1}{p-2}-\frac{p+1}{(p+1)^{2}+(\sqrt{3})^{2}}-\sqrt{3} \frac{\sqrt{3}}{(p+1)^{2}+(\sqrt{3})^{2}}\right)
\end{gathered}
$$

From here, using formulas 4,13,12 of Table 3.2, we have:

$$
\begin{aligned}
& \frac{1}{p-2} \doteqdot e^{2 t} \quad(\alpha=2) \\
& \frac{p+1}{(p+1)^{2}+(\sqrt{3})^{2}} \doteqdot e^{-t} \cos \sqrt{3} t \quad(\lambda=-1, \omega=\sqrt{3}) \\
& \frac{\sqrt{3}}{(p+1)^{2}+(\sqrt{3})^{2}} \doteqdot e^{-t} \sin \sqrt{3} t \quad(\lambda=-1, \omega=\sqrt{3}) \\
& f(t)=\frac{1}{12}\left(e^{2 t}-e^{-t} \cos \sqrt{3} t-\sqrt{3} e^{-t} \sin \sqrt{3} t\right)
\end{aligned}
$$

c) Since in this case all the roots of the denominator are real and simple (single), it is best to use the decomposition theorem - formula 15 of Table 3.1. We have:

$$
\begin{aligned}
& \overline{f_{1}}(p)=p+1 \\
& \overline{f_{2}}(p)=p(p-1)(p-2)(p-3) \equiv p^{4}-6 p^{3}+11 p^{2}-6 p \\
& \overline{f_{2}}(p)=4 p^{3}-18 p^{2}+22 p-6
\end{aligned}
$$

We find the roots of the equation $\bar{f}_{2}(p)=0: p_{1}=0, p_{2}=1, p_{3}=2$, $p_{4}=3$. Then we get:

$$
\frac{\overline{f_{1}}\left(p_{1}\right)}{\overline{f_{1}^{\prime}}\left(p_{1}\right)}=-\frac{1}{6}, \frac{\overline{f_{1}}\left(p_{2}\right)}{\overline{f_{1}^{\prime}}\left(p_{2}\right)}=1, \frac{\overline{f_{1}}\left(p_{3}\right)}{\overline{f_{1}^{\prime}}\left(p_{3}\right)}=-\frac{3}{2}, \frac{\overline{f_{1}}\left(p_{4}\right)}{\overline{f_{1}^{\prime}}\left(p_{4}\right)}=\frac{2}{3}
$$

From here, using formula 15 of Table 3.1, we find:

$$
f(t)=-\frac{1}{6}+e^{t}-\frac{3}{2} e^{2 t}+\frac{2}{3} e^{3 t}
$$