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

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.
