Answer
$\frac{2x-1}{x(x^2+1)}=\frac{-1}{x}+\frac{x+2}{x^2+1}$
Work Step by Step
$\frac{2x-1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$,
$A(x^2+1)+(Bx+C)(x)$,
$Ax^2+A+Bx^2+Cx$,
$(A+B)x^2+Cx+A=2x-1$,
$\begin{cases}
A+B=0\\
C=2\\
A=-1
\end{cases}$,
thus, substituting back into $-1+B=0, B=1$.
Therefore,
$\frac{2x-1}{x(x^2+1)}=\frac{-1}{x}+\frac{x+2}{x^2+1}$
