Alg5.1.2

Find the partial fraction decomposition of $\frac{s^2+s+1}{(s+1)(s^2-1)}\,$

Let $\frac{s^2+s+1}{(s+1)(s^2-1)}=\frac{A}{s+1}+\frac{Bs+C}{s^2-1}\,$

Simplifying on the right hand side,we get

$\frac{s^2+s+1}{(s+1)(s^2-1)}=\frac{A(s^2-1)+(Bs+C)(s+1)}{(s+1)(s^2-1)}\,$

Now

$s^2+s+1=A(s^2-1)+(Bs+C)(s+1)\,$

Let $s=1\,$

Then $1^2+1+1=A(1-1)+(Bs+C)2,2B+2C=3\,$

Let $S=2\,$ then $7=3A+6B+2C\,$

$s=-2,3=3A+2B-C\,$

Solving the equations $3A+4B=4,6A+6B=9\,$

$2B=-1,B=-\frac{1}{2},A=2,C=2\,$

Hence the partial decomposition is $\frac{2}{s+1}+\frac{-s+4}{2(s^2-1)}\,$