$\left(1+ \frac{\frac{1}{x}}{1+\frac{1}{x}}\right)\times\left(1+ \frac{\frac{1}{x}}{1- \frac{1}{x}}\right)$
Multiply both numerator and denominator of $\frac{\frac{1}{x}}{1+ \frac{1}{x}}$ by x to get $\frac{1}{x+ 1}$ and both numerator and denominator of $\frac{\frac{1}{x}}{1- \frac{1}{x}}$ to get $\frac{1}{x- 1}$. Now we have $\left(1+ \frac{1}{1+ x}\right)\times \left(1+ \frac{1}{x- 1}\right)$.Add the fractions in the left parentheses: $\left(1+ \frac{1}{1+ x}\right)= \left(\frac{1+ x}{1+ x}+ \frac{1}{1+ x}\right)= \left(\frac{x+ 2}{1+ x}\right)$
Add the fractions in the right parentheses: $\left(1+ \frac{1}{x- 1}\right)= \left(\frac{x- 1}{x- 1}+ \frac{1}{x- 1}\right)= \left(\frac{x}{x- 1}\right)$.So now the product is $\left(\frac{x+ 2}{1+ x}\right)\left(\frac{x}{x- 1}\right)= \frac{x^2+ 2x}{x^2- 1}$