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Let’s start to write the F(s) function by the following way:

f(x)=\mathcal{L}^{-1}\left\{F(s)\right\} = \mathcal{L}^{-1}\left\{\frac{s^{2}+3s+2}{(s-1)(s-2)^{3}(s-4)}\right\} = \mathcal{L}^{-1}\left\{\frac{A_{1}}{s-2}+\frac{A_{2}}{(s-2)^{2}}+\frac{A_{3}}{(s-2)^{3}}+\frac{B}{s-1}+\frac{C}{s-4}\right\}

Let’s execute the partial fraction operation as follow!

\frac{s^{2}+3s+2}{(s-1)(s-2)^{3}(s-4)}= \frac{A_{1}}{s-2}+\frac{A_{2}}{(s-2)^{2}}+\frac{A_{3}}{(s-2)^{3}}+\frac{B}{s-1}+\frac{C}{s-4} \quad \quad \bigg{/}\cdot(s-1)(s-2)^{3}(s-4)

s^{2}+3s+2 = A_{1}(s-2)^{2}(s-1)(s-4) + A_{2}(s-2)(s-3)(s-4) + A_{3}(s-1)(s-4) + B(s-2)^{3}(s-4) + C(s-2)^{3}(s-1)

Let's substitute in the equation $s=2 value! Parameter $A_3 will be then determined.

$2^{2}+3\cdot 2+2 = A_{3}(2-1)(2-4)

$4+6+2 = A_{3}\cdot 1\cdot (-2)


Next step: let’s substitute $s=1 into the same equation for obtaining parameter $B:

$1^{2}+3\cdot 1+2=B(1-2)^{3}(1-4)

$6 = B(-1)(-3)


Let’s substitute $s=4 into the same equation in order to get parameter $C!

$4+3\cdot 4+2 = C(4-2)^{3}(4-1)



Let’s write A3, B and C into the equation

$s^{2}+3s+2 = A_{1}(s-2)^{2}(s-1)(s-4)+A_{2}(s-1)(s-2)(s-4) - 6(s-1)(s-4)+2(s-2)^{3}(s-4)+\frac{5}{4}(s-2)^{3}(s-1)

where only $A_{1} and $A_{2} are unknown. $A_{1} and $A_{2} will be determined by the following way. First $s=0, then $s=-1 will be substituted in the equation obtaining the following set of two linear equations for $A_{1} and $A_{2}.

s=0 & \to & -7 = 4A_{1}-2A_{2}\\ 
s=-1 & \to & -9,25 = 3A_{1}-A_{2}
/\cdot 2
\quad \rightarrow & -7=4A_{1}-2A_{2}\\ 
\quad \rightarrow & -18,5 = 12A_{1}-2A_{2}

Let’ substitute the two equation : $11,5 = -8A_{1}

Consequently $ \left.\begin{matrix}
A_{1} & = & -1,4375\\
A_{2} & = & 0,625

Now all the coefficients of the partial fraction are known, and Laplace transformation can be executed.

$f(x)=\mathcal{L}^{-1}\left\{-\frac{1,4375}{s-2}+\frac{0,625}{(s-2)^{2}}-\frac{6}{(s-2)^{3}}+\frac{2}{s-1}+\frac{5}{4} \frac{1}{s-4}\right\}

$f(x)=\mathcal{L}^{-1}\left\{-\frac{23}{16}\frac{1}{s-2}+\frac{5}{8}\frac{1}{(s-2)^{2}}-\frac{6}{(s-2)^{3}}+\frac{2}{s-1}+\frac{5}{4} \frac{1}{s-4}\right\}



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