Answer
$\frac{x^4+x^3+x^2-x+1}{x(x^2+1)^2}=\frac{1}{x}+\frac{1}{x^2+1}+\frac{-x-2}{(x^2+1)^2}$,
Work Step by Step
$\frac{x^4+x^3+x^2-x+1}{x(x^2+1)^2}=\frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$,
$A(x^2+1)^2+(Bx+C)(x(x^2+1))+(Dx+E)(x)$,
$A(x^4+2x^2+1)+(Bx+C)(x^3+x)+(Dx+E)(x)$,
$(A+B)x^4+Cx^3+(2A+B+D)x^2+(C+E)x+A=x^4+x^3+x^2-x+1$,
$\begin{cases}
A+B=1\\
C=1\\
2A+B+D=1\\
C+E=-1\\
A=1
\end{cases}$
$A=1, 1+B=1, B=0, 1+E=-1, E=-2, 2+D=1, D=-1$
Therefore,
$\frac{x^4+x^3+x^2-x+1}{x(x^2+1)^2}=\frac{1}{x}+\frac{1}{x^2+1}+\frac{-x-2}{(x^2+1)^2}$,
