Take the simplest case
$\dfrac1{x(x-a)^2}
$.
If
$\dfrac1{x(x-a)^2}
=\dfrac{u}{x}+\dfrac{v}{(x-a)^2}
$
then
$\dfrac{u}{x}+\dfrac{v}{(x-a)^2}
=\dfrac{u(x-a)^2+vx}{x(x-a)^2}
=\dfrac{ux^2+(v-2ua)x+ua^2}{x(x-a)^2}
$
and you need to have
$u = 0, v-2ua=0, ua^2 = 1$.
This is three equations in two unknowns
and, in this case,
has no solution.
You need three parameters,
which are gotten by writing
$\dfrac1{x(x-a)^2}
=\dfrac{u}{x}+\dfrac{v}{x-a}+\dfrac{w}{(x-a)^2}
$.
This becomes
$1
=u(x-a)^2+vx(x-a)+wx
$
and this can be solved for
$u, v, w$.
The general case is
$\dfrac1{x(x-a)^m}
=\dfrac{c}{x}+\sum_{j=1}^m \dfrac{b_j}{(x-a)^j}
$
and this can be solved
for
$c$ and the $b_j$.
If we clear fractions,
this becomes
$1
= c(x-a)^m+\sum_{j=1}^m b_jx(x-a)^{m-j}
= c(x-a)^m+x\sum_{j=1}^m b_j(x-a)^{m-j}
$.
Setting $x=0$ gives
$1 = c(-a)^m$
so
$c = \dfrac1{(-a)^m}
$
and
$1
= (x-a)^m(-a)^{-m}+x\sum_{j=1}^m b_j(x-a)^{m-j}
$.
Putting $x=a$ gives
$1
= ab_m
$
so
$b_m = \dfrac1{a}
$.
The successive $b_j$
can be similarly determined.
Best Answer
$$ \frac{x+1}{x^4+x}=\frac{x+1}{x(x+1)(x^2-x+1)}=\frac{1}{x(x^2-x+1)}=\frac{1}{x}-\frac{x+1}{x^2-x+1} $$