Answer
$\frac{2x-1}{x(x^2+1)^2}= \frac{A}{x} + \frac{Bx+C}{(x^2+1)}+ \frac{Dx+E}{(x^2+1)^2}$
Work Step by Step
Rewrite the fraction using partial fractions
Because the denominator is already in simplest terms, begin the decomposition.
Note that for quadratic terms an $Ax+B$ (or other constants for A and B) must be used
$\frac{2x-1}{x(x^2+1)^2}= \frac{A}{x} + \frac{Bx+C}{(x^2+1)}+ \frac{Dx+E}{(x^2+1)^2}$
