Solution:

$\frac{x^{2}+23 x}{(x+3)\left(x^{2}+1\right)}=\frac{A}{x+3}+\frac{B x+C}{x^{2}+1}$

$\therefore x^{2}+23 x=\left(x^{2}+1\right) A+(x+3)(B x+C)$

Put $x=-3$

$\therefore(-3)^{2}+23(-3)=\left((-3)^{2}+1\right) A$

$\therefore-60=10 A$

(\therefore A=-6$ Put $x=0$ $\therefore 0=(1) A+(3)(0+C)$ $\therefore 0=-6+3 C$ $\therefore C=2$ Put $x=1$ $\therefore 24=2(-6)+4 B+4(2)$ $\therefore B=7$ $\therefore B=7$ $\therefore \frac{x^{2}+23 x}{(x+3)\left(x^{2}+1\right)}=\frac{-6}{x+3}+\frac{7 x+2}{x^{2}+1}$