Simplifying $\frac{\;\frac{x}{1-x}+\frac{1+x}{x}\;}{\frac{1-x}{x}+\frac{x}{1+x}}$ 
$$\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}$$

I can simplify this using the slow way ---adding the numerator and denominator, and then dividing--- but my teacher told me there is another way. Any help?
 A: Multiply the numerator and denominator by $x(1+x)(1-x)$. This clears away all the "inner" denominators, leaving $${xx(1+x)+(1+x)(1+x)(1-x)\over (1-x)(1-x)(1+x)+xx(1-x)}={(1+x)\over (1-x)}{x^2+1-x^2\over x^2+1-x^2}={1+x\over 1-x}.$$
A: Multiply both the numerator and the denominator by $x(1-x)(1+x)$ and then use the fact that $(1-x)(1+x)=1-x^2$ to see that
\begin{align}
\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}&=\left(\frac{\;\;\dfrac{x}{1-x}+\dfrac{1+x}{x}\;\;}{\dfrac{1-x}{x}+\dfrac{x}{1+x}}\right)\frac{x(1-x)(1+x)}{x(1-x)(1+x)}
\\&=\left(\frac{x^2+(1+x)(1-x)}{(1-x)(1+x)+x^2}\right)\frac{1+x}{1-x}
\\&=\left(\frac{1}{1}\right)\frac{1+x}{1-x}
\\&=\frac{1+x}{1-x}.
\end{align}
A: If you note that the expression, which is of form $$\cfrac{a+b}{\frac 1a +\frac 1b}$$ simplifies easily to $ab$ (multiply through by $\frac {ab}{ab}$) you can avoid a lot of potentially confusing/error inducing algebra, and easily get the result that others have given.
It is sometimes useful to identify the form of an expression and simplify that, especially when it looks likely to get confusing. Sometimes it is a distraction, but here it would be a quick try and you could revert to a more hands-on approach if it doesn't work.
A: Let me try. 
$$\frac{\frac{x}{1-x}+\frac{1+x}{x}}{\frac{1-x}{x}+\frac{x}{1+x}} = \frac{\frac{1}{1-x}-1 + \frac{1}{x} + 1}{\frac{1}{x}-1 + 1-\frac{1}{1+x}} = \frac{\frac{1}{1-x} + \frac{1}{x}}{\frac{1}{x} - \frac{1}{1+x}} = \frac{\frac{1}{x(1-x)}}{\frac{1}{x(1+x)}} = \frac{1+x}{1-x}$$
