Solve Diff Eq Problem (s+1)/s*(s^2+s+1): Get Expert Help

[bboy]

(s+1)/s*(s^2+s+1)
find laplace transform..
i am confused how to solve this problem
i did numerator separation and i did A/s + (Bs+c)/(s^2+s+1) but still doenst work..so experts..i do really need ur help..thnx in advance

 

[benorin]

$$\frac{s+1}{s(s^2+s+1)}=\frac{1}{s}-\frac{s}{s^2+s+1}=\frac{1}{s}-\frac{s\frac{1}{2}}{\left(s+\frac{1}{2}\right)^2+\frac{3}{4}}+\frac{\frac{1}{2}}{\left(s+\frac{1}{2}\right)^2+\frac{3}{4}}=\frac{1}{s}-\frac{s+\frac{1}{2}}{\left(s+\frac{1}{2}\right)^2+\frac{3}{4}}+\frac{1}{\sqrt{3}}\cdot\frac{\frac{\sqrt{3}}{2}}{\left(s+\frac{1}{2}\right)^2+\frac{3}{4}}$$​

so the inverse transformation is

$$\mathcal{L}^{-1}\left\{\frac{s+1}{s(s^2+s+1)}\right\}=\mathcal{L}^{-1}\left\{\frac{1}{s}-\frac{s+\frac{1}{2}}{\left(s+\frac{1}{2}\right)^2+\frac{3}{4}}+\frac{1}{\sqrt{3}}\cdot\frac{\frac{\sqrt{3}}{2}}{\left(s+\frac{1}{2}\right)^2+\frac{3}{4}}\right\}=u(t)-u(t)e^{-\frac{1}{2}t}\cos\left(\frac{\sqrt{3}}{2}t}\right)+u(t)e^{-\frac{1}{2}t}\sin\left(\frac{\sqrt{3}}{2}t}\right)$$​

where $$u(t)$$ is the unit step function and I got the inverse transforms from here.