<s>[INST] How to show that $\frac{x^2}{x-1}$ simplifies to $x + \frac{1}{x-1} +1$ How does $\frac{x^2}{(x-1)}$ simplify to $x + \frac{1}{x-1} +1$?
The second expression would be much easier to work with, but I cant figure out how to get there.
Thanks
 [/INST] A general $^1$ method is to perform the polynomial long division algorithm in the following form or another equivalent one:

to get
$$
x^{2}=(x-1)(x+1)+1.
$$
Divide both sides by $x-1$:
$$\begin{eqnarray*}
\frac{x^{2}}{x-1} &=&\frac{(x-1)(x+1)+1}{x-1}=\frac{(x-1)(x+1)}{x-1}+\frac{1}{x-1} \\
&=&\frac{x+1}{1}+\frac{1}{x-1}=x+1+\frac{1}{x-1}=x+\frac{1}{x-1}+1
\end{eqnarray*}
$$
--
$^1$ The very same method can be applied to find (or show) that
$$\frac{x^{4}-2x^{3}-6x^{2}+12x+15}{x^{3}+x^{2}-4x-4}=x-3+\frac{
x^{2}+4x+3}{x^{3}+x^{2}-4x-4};$$
In this case the polynomial long division algorithm corresponds to the following computation (or another equivalent one):

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