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