Answer
$\frac{2x-1}{x(x^2+1)}=\frac{-1}{x}+\frac{x+2}{x^2+1}$
Work Step by Step
$\frac{2x-1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}$,
$A(x^2+1)+(Bx+C)x$,
$Ax^2+A+Bx^2+Cx$,
$(A+B)x^2+Cx+A=2x-1$,
thus, $C=2, A=-1, B=1$
Therefore,
$\frac{2x-1}{x(x^2+1)}=\frac{-1}{x}+\frac{x+2}{x^2+1}$
