Solution:

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

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

Put x=1

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

$\therefore A=\frac{3}{2}$

Put x=0

$\therefore 2(0)+1=(0+1) A+(0-1)(B(0)+C)$

$\therefore 1=A-C$

$\therefore 1=\frac{3}{2}-C$

$\therefore C=\frac{1}{2}$

Put x=-1

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

$\therefore-1=2 A+2 B-2 C$

$\therefore-1=2\left(\frac{3}{2}\right)+2 B-2\left(\frac{1}{2}\right)$

$\therefore-1=3+2 B-1$

$\therefore B=-\frac{3}{2}$

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