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$\frac{(s^2 +s +1)}{(s^2+4s+3)(s+1)}$

the answer has to be in a $\frac{A}{(s+1)} + \frac{Bs+C}{(s^2+4s+3)}$ form. However i tried to solve it this way but end up with that there is no solution for this problem. For instance i get: $A=1+A$ to solve for A which is false. Please help me i spent a lot of time solving this question with different ways and have no ideas left.

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  • $\begingroup$ Please add steps how you got $A=A+1.$ Otherwise no one can help... $\endgroup$ – coffeemath Nov 17 '18 at 12:18
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There must be some mistake in the question as $s^2+4s+3=(s+1)(s+3)$

Using Partial Fraction Decomposition,

$\dfrac{s^2+s+1}{(s+1)(s^2+4s+3)}=\dfrac A{s+1}+\dfrac B{(s+1)^2}+\dfrac C{s+3}$

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$$\frac{{{s}^{2}}+s+1}{\left( s+1\right) \, \left( {{s}^{2}}+4 s+3\right) }=\frac{7}{4 \left( s+3\right) }-\frac{3}{4 \left( s+1\right) }+\frac{1}{2 {{\left( s+1\right) }^{2}}}$$

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Note that $$\frac {s^2+s+1}{(s+1)(s^2+4s+3)}=\frac {A}{s+1}+ \frac {B}{(s+1)^2}+\frac {C}{s+3}$$

I have found $A=-3/4$, $B=1/2$ and $C=7/4$

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  • $\begingroup$ nice! thanks! but how can i transform it to this form? $\frac{A}{(s+1)}+\frac{(Bs+C)}{s^2+4s+3}$ $\endgroup$ – David Nov 17 '18 at 12:38
  • $\begingroup$ @Mohammad Riazi-Kermani Your answer is wrong $\endgroup$ – Aleksas Domarkas Nov 17 '18 at 15:05
  • $\begingroup$ @Alek. Fixed it $\endgroup$ – Mohammad Riazi-Kermani Nov 17 '18 at 15:15
  • $\begingroup$ got it now! thanks so much @MohammadRiazi-Kermani $\endgroup$ – David Nov 17 '18 at 15:27
  • $\begingroup$ Thanks for your attention. $\endgroup$ – Mohammad Riazi-Kermani Nov 17 '18 at 17:13

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