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
 A: $\textbf{Hint:}$ Do you know about polynomial long divison?
A simpler way of dealing with this problem is noticing that for all $x\neq 1$, 
$$\frac{x^2}{x-1}=\frac{x^2-1+1}{x-1}=\frac{x^2-1}{x-1}+\frac{1}{x-1}=\frac{(x-1)(x+1)}{x-1}+\frac{1}{x-1}=x+1+\frac{1}{x-1}.$$
It's not always this easy, though. So you should learn polynomial long divison.
A: 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):

A: We have that
$$
x^2=(x-1+1)^2=(x-1)^2+1+2(x-1)
$$
and so for $x\neq 1$:
$$
\frac{x^2}{x-1}=\frac{(x-1)^2+1+2(x-1)}{x-1}=x-1+\frac{1}{x-1}+2=x+1+\frac{1}{x-1}.
$$
A: Very clever trick: If you have to show that two expressions are equivalent, you work backwards. $$\begin{align}=& x +\frac{1}{x-1} + 1 \\ \\ \\ =& \frac{x^2 - x}{x-1} +\frac{1}{x - 1} + \frac{x-1}{x-1} \\ \\ \\ =& \frac{x^2 - x + 1 + x - 1}{x - 1} \\ \\ \\ =&\frac{x^2 }{x - 1}\end{align}$$Now, write the steps backwards (if you're going to your grandmommy's place, you turn backwards and then you again turn backwards, you're on the right way!) and act like a know-it-all.
$$\begin{align}=&\frac{x^2 }{x - 1} \\ \\ \\ =& \frac{x^2 - x}{x-1} +\frac{1}{x - 1} + \frac{x-1}{x-1} \\ \\ \\ =& x +\frac{1}{x-1} + 1 \end{align}$$
Q.E.Doodly dee!
This trick works and you can impress your friends with such elegant proofs produced by this trick.
A: $$\frac{x^2}{x-1}=\frac{x^2-x+x-1+1}{x-1} = \frac{x(x-1) + (x-1)+1}{x-1}=\frac{x(x-1)}{x-1}+\frac{x-1}{x-1}+\frac{1}{x-1}$$
A: Hint: An alternative method would be to compute: $$\frac{x^2}{x-1}-\left(x+1+\frac1{x-1}\right)$$
And then conclude that since $$\frac{x^2}{x-1}-\left(x+1+\frac1{x-1}\right)=0$$
Then $$\frac{x^2}{x-1}=x+1+\frac1{x-1}.$$
