I'm just looking for some overall clarification for the following cases. Now, to the extent of my knowledge, the following examples of partial fractions would be split up in the following way: \begin{align} \frac{1}{x^2+3x-4}&=\frac{1}{\left(x+4\right)\left(x-1\right)}=\frac{A}{x+4}+\frac{B}{x-1},\tag{1}\\ \frac{1}{x^3+x^2}&=\frac{1}{x^2\left(x+1\right)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1},\tag{2}\\ \frac{1}{x^2\left(x+1\right)^3}&=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}+\frac{D}{\left(x+1\right)^2}+\frac{E}{\left(x+1\right)^3},\tag{3}\\ \frac{1}{x^2\left(x^2+x+1\right)}&=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+x+1},\tag{4}\\ \frac{1}{x^2\left(x^3+x^2+x+1\right)^2}&=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx^2+Dx+E}{x^3+x^2+x+1}+\frac{Fx^2+Gx+H}{\left(x^3+x^2+x+1\right)^2},\tag{5} \end{align} have I made a mistake anywhere?
EDIT: I should have factored $\left(5\right)$ further to get \begin{align} \frac{1}{x^2\left(\left(x+1\right)\left(x^2+1\right)\right)^2}. \end{align}
$$ x^3+x^2+x+1 = x^2(x+1) + 1(x+1) = (x^2+1)(x+1) $$ So $$ \frac{\cdots\cdots\cdots}{x^3+x^2+x+1} = \frac{Cx+D}{x^2+1} + \frac{E}{x+1}. $$