Answer
$\frac{x^2+x-2}{x^2+x}$
Work Step by Step
Step 1. Perform operations to the numerator: $2+\frac{2}{1+x}=\frac{2(1+x)+2}{1+x}=\frac{4+2x}{1+x}$
Step 2. Perform operations to the denominator: $2-\frac{2}{1-x}=\frac{2(1-x)-2}{1-x}=\frac{-2x}{1-x}$
Step 3. Perform the division: $\frac{2+\frac{2}{1+x}}{2-\frac{2}{1-x}}=\frac{\frac{4+2x}{1+x}}{\frac{-2x}{1-x}}=\frac{4+2x}{1+x}\cdot \frac{1-x}{-2x}=\frac{(x+2)(x-1)}{x(x+1)}=\frac{x^2+x-2}{x^2+x}$
