Simplify [1/(x-1) + 1/(x²-1)] / [x-2/(x+1)] Simplify: $$\frac{\frac{1}{x-1} + \frac{1}{x^2-1}}{x-\frac 2{x + 1}}$$ 
This is what I did.
Step 1: I expanded $x^2-1$ into: $(x-1)(x+1)$. And got: $\frac{x+1}{(x-1)(x+1)} + \frac{1}{(x-1)(x+1)}$
Step 2: I calculated it into: $\frac{x+2}{(x-1)(x+1)}$
Step 3: I multiplied $x-\frac{2}{x+1}$ by $(x-1)$ as following and I think this part might be wrong:  


*

*$x(x-1) = x^2-x$. Times $x+1$ cause that's the denominator =   

*$x^3+x^2-x^2-x = x^3-x$.   

*After this I added the $+ 2$

*$\frac{x^3-x+2}{(x-1)(x+1)}$


Step 4: I canceled out the denominator $(x-1)(x+1)$ on both sides.
Step 5: And I'm left with: $\frac{x+2}{x^3-x+2}$
Step 6: Removed $(x+2)$ from both sides I got my UN-correct answer: $\frac{1}{x^3}$
Please help  me. What am I doing wrong?

First recall that
\[ \frac{a}{b} \pm \frac{c}{d} =\frac{ad \pm cb}{bd} \]
And 
\[ \frac{\frac{a}{b}}{\frac{c}{d}} =\frac{ad}{bc} \]
Now just simplify, no fancy fractions needed:
\[
\frac{\frac{1}{x-1}+\frac{1}{x^2-1}}{x-\frac{2}{x+1}}
= \frac{\frac{(x^2-1)+(x-1)}{(x-1)(x^2-1)}}{\frac{x(x+1)-2}{x+1}}
= \frac{\frac{(x-1)(x+1)+(x-1)}{(x-1)^2(x+1)}}{\frac{(x+2)(x-1)}{x+1}}
= \frac{\frac{(x-1)(x+2)}{(x-1)^2(x+1)}}{\frac{(x+2)(x-1)}{x+1}}
= \frac{(x-1)(x+2)(x+1)}{(x-1)^3(x+1)(x+2)}
= \frac{1}{(x-1)^2}
\]
I attempted to be as clear as possible. If you'd like me to elaborate further, just let me know.
