We have
\begin{equation} F(s)=\frac{s^2+1}{(s+1)(s-1)} \end{equation}
which I want to use Heavisde method to find the fractions.
We start
\begin{equation} F(s)=\frac{s^2+1}{(s+1)(s-1)}=\frac{A}{(s+1)}+\frac{B}{(s-1)} \end{equation}
By Heaviside method, we multiply both sides by the denumerator of A, ($s+1$)
\begin{equation} (s+1)\frac{s^2+1}{(s-1)}=A+\frac{B(s+1)}{(s-1)} \end{equation}
We insert for the pole of the A-fraction, for s, that is $s=-1$ and we obtain
\begin{equation} (-1+1)\frac{s^2+1}{(s-1)}=A+\frac{B(0)}{(-2)} \end{equation}
which gives the incorrect results, that $A=0$
What should I do here to prepare the fractions correctly for the Heaviside method to be applied?
Thanks
Update, I devise the method suggested below in the accepted solution, and try the Heaviside method on that.
\begin{equation} F(s)=\frac{s^2+1}{(s+1)(s-1)}=\frac{(s^2-1)+2}{(s+1)(s-1)} \end{equation}
\begin{equation} F(s)=\frac{(s^2-1)+2}{(s+1)(s-1)}=\frac{(s+1)(s-1)+2}{(s+1)(s-1)} \end{equation}
\begin{equation} F(s)=\frac{(s+1)(s-1)+2}{(s+1)(s-1)}=\frac{(s+1)(s-1)}{(s+1)(s-1)}+\frac{2}{{(s+1)(s-1)}} \end{equation}
\begin{equation} F(s)=1+\frac{2}{{(s+1)(s-1)}} \end{equation}
So from here we treat the last fraction with the Heaviside method:
\begin{equation} F(s)=\frac{2}{{(s+1)(s-1)}}=\frac{A}{(s+1)}+\frac{B}{(s-1)} \end{equation}
Find A by multiplying with $(s+1)$ on both sides and obtain
\begin{equation} (s-1)=A+\frac{B(s+1)}{(s-1)} \end{equation}
Insert for the pole of the A-fraction, $s=-1$
\begin{equation} (-1-1)=A+\frac{B(-1+1)}{(s-1)} \end{equation}
This yields that $A=-2$. We do the same for B,
\begin{equation} (s+1)=\frac{A(s-1)}{(s+1)}+B \end{equation}
Insert for $s=1$, and get that $B=2$, hence:
\begin{equation} F(s)=1+\frac{2}{(s-1)}-\frac{2}{(s+1)} \end{equation}
Inverse transform of F(s):
\begin{equation} \mathscr{L}^{-1}\{1+\frac{2}{(s-1)}-\frac{2}{(s+1)}\}=\delta(t)+2e^{t}-2e^{-t} \end{equation}
Here is another way to approach it for the sake of curiosity.
To start with, notice that \begin{align*} \frac{s^{2} + 1}{(s-1)(s+1)} & = \frac{(s^{2} - 1) + 2}{s^{2} - 1}\\\\ & = 1 + \frac{2}{s^{2} - 1} \end{align*}
Once you have done so, observe that \begin{align*} \frac{2}{s^{2} - 1} & = \frac{(s + 1) - (s - 1)}{(s-1)(s+1)}\\\\\ & = \frac{1}{s - 1} - \frac{1}{s + 1} \end{align*}
Gathering all the previous results, it results that \begin{align*} \frac{s^{2} + 1}{(s-1)(s+1)} = 1 + \frac{1}{s - 1} - \frac{1}{s + 1} \end{align*}
Hopefully this helps !