\(\displaystyle \frac{2x-3}{x(x^2+1)} \Rightarrow 2x-3 = \frac{A}{x} + \frac{Bx+C}{x^2+1} \) The question is, how does the numerator on \(\displaystyle x^2 + 1 \) equal \(\displaystyle Bx + C \) ?

You need to combine the two fractions such that there will be no \(\displaystyle x^2\) term in the resulting numerator

\(\displaystyle \frac{A}{x}+\frac{Bx+C}{x^2+1}=\frac{x^2+1}{x^2+1}\ \frac{A}{x}+\frac{x}{x}\ \frac{Bx+C}{x^2+1}\)

\(\displaystyle =\frac{Ax^2+A+Bx^2+Cx}{x\left(x^2+1\right)}\)

This allows the \(\displaystyle x^2\) terms to cancel as \(\displaystyle B=-A,\)

Also \(\displaystyle C=2,\ A=-3\)