# How can we take the inverse Laplace transform of f_1(s)X(s)+f_2(s)(e^(i phi)X(s−i alpha_1)+e^(-i phi) X(s+i alpha_1))=f_0(s)

How can we take the inverse Laplace transform of
${f}_{1}\left(s\right)X\left(s\right)+{f}_{2}\left(s\right)\left({e}^{i\varphi }X\left(s-i{\alpha }_{1}\right)+{e}^{-i\varphi }X\left(s+i{\alpha }_{1}\right)\right)={f}_{0}\left(s\right)$
Where ${f}_{1}\left(s\right)$ is in the form of $\frac{{f}^{2}\left(s\right)}{{f}^{3}\left(s\right)}$, ${f}_{2}\left(s\right)={k}_{0}s$, and ${f}_{0}\left(s\right)$ is in the form of ${k}_{1}+\frac{{f}^{4}\left(s\right)}{s\left({s}^{2}+{\alpha }_{0}^{2}\right){f}^{2}\left(s\right)}$
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Farbwolkenw
Hint.
Anti-transforming
${e}^{i\varphi }X\left(s-i{\alpha }_{1}\right)+{e}^{-i\varphi }X\left(s+i{\alpha }_{1}\right)={G}_{0}\left(s\right)+{G}_{1}\left(s\right)X\left(s\right)$
we get at
$\mathrm{cos}\left({\alpha }_{1}\phantom{\rule{thinmathspace}{0ex}}t+\varphi \right)\phantom{\rule{thinmathspace}{0ex}}{x}_{3}\left(t\right)={\int }_{0}^{t}{g}_{1}\left(\tau \right){x}_{3}\left(t-\tau \right)d\tau +{g}_{0}\left(t\right)$
an integral equation for ${x}_{3}\left(t\right)$