help in Laplace and partial fractions Can any one teach me how to solve C2.(a) and (b) step by step? 

C2. (a) Resolve $\frac{1}{s^2(s^2+s+1)}$ into partial fractions of the form $\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+s+1}$. 
Hence, resolve $\frac{1}{s(s^2+s+1)}$ into its partial fractions.
C2. (b) Find
$$\mathcal{L}^{-1}\left\{\frac{s}{s^2+s+1}\right\}.$$
 A: (a) The function
$$ f(z)=\frac{1}{z^2(z^2+z+1)}$$
has a double pole in $z=0$ and a simple pole in $z=e^{\pm 2\pi i/3}$, since $z^2+z+1=\frac{z^3-1}{z-1}$. This gives:
$$ f(z)=\frac{A}{z^2}+\frac{B}{z}+\frac{C}{z-\omega}+\frac{D}{z-\omega^2}+g(z)$$
where $g(z)$ is a holomorphic function. By computing:
$$\lim_{z\to 0}z^2 f(z)=1,\quad \lim_{z\to\omega}f(z)(z-\omega)=\frac{1}{\sqrt{3}}e^{i\pi/6},\quad \lim_{z\to\omega}f(z)(z-\omega^2)=\frac{1}{\sqrt{3}}e^{-i\pi/6}$$
we have:
$$ f(z)=\frac{1}{z^2}+\frac{z}{z^2+z+1}+\frac{B}{z}+g(z) $$
where $B=-1$ since the sum of the residues must be zero. 
Rearranging and checking that $g(z)=0$, we get:
$$ f(z) = \frac{1}{z^2}-\frac{1}{z}+\frac{e^{i\pi/6}}{\sqrt{3}(z-\omega)}+\frac{e^{-i\pi/6}}{\sqrt{3}(z-\omega^2)}=\frac{1}{z^2}-\frac{1}{z}+\frac{z}{z^2+z+1}.$$
Multypling by $z$ and rearranging we have:
$$ \frac{1}{z(z^2+z+1)}=\frac{1}{z}+\frac{-3+i\sqrt{3}}{6(z-\omega)}+\frac{-3-i\sqrt{3}}{6(z-\omega^2)}=\frac{1}{z}-\frac{z+1}{z^2+z+1}.$$
(b) Since 
$$\frac{z}{z^2+z+1}=\frac{e^{i\pi/6}}{\sqrt{3}(z-\omega)}+\frac{e^{-i\pi/6}}{\sqrt{3}(z+\omega)}$$
and
$$\mathcal{L}^{-1}\left(\frac{1}{z-\xi}\right)=e^{s\xi},$$
by linearity it follows that
$$\color{red}{\mathcal{L}^{-1}\left(\frac{z}{z^2+z+1}\right)} = \frac{1}{\sqrt{3}}\left(e^{i\pi/6}e^{\omega s}+e^{-i\pi/6}e^{\omega^2 s}\right)=\color{red}{e^{-s/2}\left(\cos\frac{\sqrt{3}\,s}{2}-\frac{1}{\sqrt{3}}\sin\frac{\sqrt{3}\,s}{2}\right)}$$
