# Partial fraction for complex roots using the second order polynomial

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

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.

• Please add steps how you got $A=A+1.$ Otherwise no one can help... – coffeemath Nov 17 '18 at 12:18

There must be some mistake in the question as $$s^2+4s+3=(s+1)(s+3)$$

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

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

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 $$C=7/4$$

• nice! thanks! but how can i transform it to this form? $\frac{A}{(s+1)}+\frac{(Bs+C)}{s^2+4s+3}$ – David Nov 17 '18 at 12:38