What is the inverse Laplace transform of the equation? $F(s)=\frac{1}{s\left( s^2 +s+1\right)}$?

Question

$F(s)=\frac{1}{s\left( s^2 +s+1\right)}$

Can you help me? Is there someone to explain step by step?

Answer 1

\begin{gather*} \frac{1}{s\left( s^{2} +s+1\right)} =\frac{s^{2} +s+1-(s^{2} +s)}{s\left( s^{2} +s+1\right)} =\frac{1}{s} -\frac{s+1}{s^{2} +s+1}\\ =\frac{1}{s} -\frac{s+1}{\left( s+\frac{1}{2}\right)^{2} +\left(\frac{\sqrt{3}}{2}\right)^{2}}\\ =\frac{1}{s} -\frac{s+\frac{1}{2}}{\left( s+\frac{1}{2}\right)^{2} +\left(\frac{\sqrt{3}}{2}\right)^{2}} -\frac{1}{2}\frac{1}{\left( s+\frac{1}{2}\right)^{2} +\left(\frac{\sqrt{3}}{2}\right)^{2}}\\ L^{-1}\left(\frac{1}{s}\right) =1\\ L^{-1}\left(\frac{s+a}{( s+a)^{2} +b^{2}}\right) =e^{-at}\cos bt\\ L^{-1}\left(\frac{a}{( s+a)^{2} +b^{2}}\right) =e^{-at}\sin bt \end{gather*} Can you use the information provided to get at the final answer?

(Hint: Do some rearranging.)