Where does the x come from when you decompose non-linear denominators? I have done >4h of research with different text books and I'm still stucked. 
I also have looked at enter link description here
But still I don't get it, where does this x at Bx+C come from?: 
$$ \frac{3}{(x+1)(x²+4)} = \frac{A}{(x+1)}+\frac{Bx+C}{(x²+4)} $$
The thing I am struggling with is: I don't have any clue about the connection between the denominator and the numerator on the right hand side. 
All I know is that the numerators of the right hand side should give, when multiplied by the denominators 3 and that the product of the denominators should be (x+1)(x²+4). So how do we know that there has to be an x in Bx+C ? 
My textbook says: "Now if we are dividing by x²+4 the remainder can have xs in, as well as numbers" And I don't know what "we" are dividing to get a remainder like that ? 3/(x²+4) ? 


Maybe someone could give a really really really easy step by step example, how you receive that Bx term rather than why it is logical to be there?


Like my textbook mentioned, there's a way to simply divide "something" and receive a remainder of that form .

As mentioned in one of my comments, Hagen von Eitzen's answer to the linked question may provide a satisfactory answer.
If you want to see an explicit computation, consider a modification of your example:
$$
\frac{10x^2+5}{(x+1)(x^2+4)}.
$$
Noting that $x+1$ and $x^2+4$ have no non-constant polynomial factor in common, we can, using the Euclidean algorithm, write $1=u(x)(x^2+4)+v(x)(x+1)$.  To find $u(x)$ and $v(x)$ note that $x^2+4$ divided by $x+1$ gives quotient $x-1$ and remainder $5$.  So $5=(x^2+4)-(x-1)(x+1)$, implying that $u(x)=\frac{1}{5}$ and $v(x)=-\frac{1}{5}(x-1)$.  We may now write the original rational function as
$$
\begin{aligned}
\frac{(2x^2+1)\cdot5}{(x+1)(x^2+4)}&=\frac{(2x^2+1)[(x^2+4)-(x-1)(x+1)]}{(x+1)(x^2+4)}=\frac{2x^2+1}{x+1}-\frac{(2x^2+1)(x-1)}{x^2+4}\\
&=\frac{2x^2+1}{x+1}-\frac{2x^3-2x^2+x-1}{x^2+4}.
\end{aligned}
$$
By polynomial division we may write both terms in the form
$$
\frac{\text{numerator}}{\text{denominator}}=\text{quotient}+\frac{\text{remainder}}{\text{denominator}}
$$
Since the numerator of the original rational function has lower degree than does the denominator, it is clear that the quotient terms are going to have to cancel, leaving only the remainder terms.  Indeed
$$
\frac{2x^2+1}{x+1}=2x-2+\frac{3}{x+1}
$$
and
$$
\frac{2x^3-2x^2+x-1}{x^2+4}=2x-2+\frac{-7x+7}{x^2+4},
$$
leaving
$$
\frac{10x^2+5}{(x+1)(x^2+4)}=\frac{3}{x+1}+\frac{7x-7}{x^2+4}.
$$
