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$$\frac{1}{p-a}\cdot \frac{p+ia}{p-ia}=\frac{p+ia}{p^2+a^2}=\frac{p}{p^2+a^2}+i\frac{a}{p^2+a^2}$$

\begin{align*}
p=0\Longrightarrow -1=-A\Longrightarrow \ \ A=1 \\
p=1\Longrightarrow 4=2B\Longrightarrow \ \ B=2 \\
p=-1, p=2 &\Longrightarrow  C=-3,\ \ D=-2 \\
 & \Longrightarrow F(p)=\frac{1}{p}+\frac{2}{p-1}-\frac{3p+2}{s^2+1} \\
 & \Longrightarrow L^{-1}[F(p)]=L^{-1}[\frac{1}{p}]+L^{-1}[\frac{2}{p-1}]-L^{-1}[\frac{3p}{p^2+1}]-L^{-1}[\frac{2}{p^2+1}] \\
 & =1+3e^t-3cost-2sint
\end{align*}

\end{document}