# Let $$\displaystyle{x}^{{\ast}}{\left({s}\right)}={\left({\frac{{{1}}}{{{\left({s}+\mu_{{1}}+\mu_{{2}}\right)}{\left({s}+\hat{{\lambda}}_{{2}}\right)}{\left({s}+\lambda_{{1}}+\lambda_{{2}}\right)}}}}\right)}$$ be the laplace transform in question, where

Let
${x}^{\ast }\left(s\right)=\left(\frac{1}{\left(s+{\mu }_{1}+{\mu }_{2}\right)\left(s+{\stackrel{^}{\lambda }}_{2}\right)\left(s+{\lambda }_{1}+{\lambda }_{2}\right)}\right)$
be the laplace transform in question, where ${\mu }_{1},{\mu }_{2},{\lambda }_{1},{\lambda }_{2},{\stackrel{^}{\lambda }}_{2}$ are just real parameters.
How can i get its inverse by means of convolution, $x\left(t\right)={\mathcal{L}}^{-1}\left({x}^{\ast }\left(s\right)\right)$
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abangan85s0

You right, first thing here is a convolution
$x\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{\left(s+{\mu }_{1}+{\mu }_{2}\right)\left(s+{\stackrel{^}{\lambda }}_{2}\right)\left(s+{\lambda }_{1}+{\lambda }_{2}\right)}\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\mu }_{1}+{\mu }_{2}}\right)*{\mathcal{L}}^{-1}\left(\frac{1}{s+{\stackrel{^}{\lambda }}_{2}}\right)*{\mathcal{L}}^{-1}\left(\frac{1}{s+{\lambda }_{1}+{\lambda }_{2}}\right)=f\left(t\right)*g\left(t\right)*w\left(t\right)$
where $f\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\mu }_{1}+{\mu }_{2}}\right)$ , $g\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\stackrel{^}{\lambda }}_{2}}\right)$ and $w\left(t\right)={\mathcal{L}}^{-1}\left(\frac{1}{s+{\lambda }_{1}+{\lambda }_{2}}\right)$
Just to note, since convolution is associative and commutative, you can evaluate it in any order (which is why no brackets were needed above),
$x\left(t\right)=\left(f\left(t\right)\cdot g\left(t\right)\right)\cdot w\left(t\right)=f\left(t\right)\cdot \left(g\left(t\right)\cdot w\left(t\right)\right)=\left(g\left(t\right)\cdot f\left(t\right)\right)\cdot w\left(t\right)=f\left(t\right)\cdot \left(w\left(t\right)\cdot g\left(t\right)\right)=\left(f\left(t\right)\cdot w\left(t\right)\right)\cdot g\left(t\right)\right)$
Now use the formula $\left(f\cdot g\right)\left(t\right)={\int }_{0}^{t}f\left(\tau \right)g\left(t-\tau \right)d\tau$ and finish the transform.

###### Not exactly what you’re looking for?
cab65699m
By the residue theorem:
$\frac{1}{\left(x-a\right)\left(x-b\right)\left(x-c\right)}=\frac{1}{\left(a-b\right)\left(a-c\right)\left(x-a\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)\left(x-b\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)\left(x-c\right)}$
and ${\mathcal{L}}^{-1}\left(\frac{1}{x-c}\right)={e}^{cs}$