Solution:

$\frac{x^{2}- x + 3}{(x - 2)(x^{2}+1)} = \frac{A}{x-2} + \frac{Bx+C}{x^{2}+1}$

$x^{2} - x +3 = (x^{2}+1)A+(x-2)(Bx+C)$

Put x=2

5 = 5A

A = 1

put x = 0

3 = A - 2C

C = 1

put x = 1

3 = 2A + (-1)(B+C)

3 = 2 - B + 1

B = 0

$\frac{x^{2}- x + 3}{(x - 2)(x^{2}+1)} = \frac{1}{x-2} + \frac{0x-1}{x^{2}+1}$

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