How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$?

Question

How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$ ?

$A = \left.\frac{s^2-s}{s^2+1} \right\vert_{s=-1} = \frac{1-(-1)}{1+1}=1$

Answer 1

We have $$ \frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1} \tag1 $$ Multiplying $(1)$ by $s$ and making $s \to \infty$ gives $$ 1=A+B $$ from which $\color{red}{B=0}$.

Making $s=0$ in $(1)$ gives $$ 0=A+C $$ from which $\color{red}{C=-1}$.

Answer 2

Cross-multiplying on the right hand side gives

$As^{2} + A + Bs^{2} + Bs + Cs + C$ on the numerator. Plug A = 1

$(1 + B)s^{2} + (B + C)s + (1 + C) = s^{2} - s$. Equate the coefficients

$1 + B = 1 \rightarrow B = 0$

$B + C = -1 \rightarrow C = -1$

Answer 3

Notice, $$\frac{s-1}{s+1}\frac{s}{s^2+1}=\frac{A}{s+1}+\frac{Bs+C}{s^2+1}$$ $$s^2-s=(A+B)s^2+(B+C)s+(A+C)$$ comparing the corresponding coefficients on both the sides, one should get $$A+B=1\tag 1$$ $$B+C=-1\tag 2$$ $$A+C=0\tag 3$$ on solving (1), (2) & (3), one can easily get $\color{red}{A=1}, \color{blue}{B=0}, \color{red}{C=-1}$