Answer
$\dfrac{4x}{2x^2+x+3}$
Work Step by Step
Let's note:
$$E=\dfrac{\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}}{\dfrac{x-1}{x+1}+\dfrac{x+2}{x-1}}.$$
Multiply both numerator and denominator by $(x-1)(x+1)$ and simplify:
$$\begin{align*}
E&=\dfrac{(x-1)(x+1)}{(x-1)(x+1)}\cdot \dfrac{\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}}{\dfrac{x-1}{x+1}+\dfrac{x+2}{x-1}}\\
&=\dfrac{(x-1)(x+1)\dfrac{x+1}{x-1}-(x-1)(x+1)\dfrac{x-1}{x+1}}{(x-1)(x+1)\dfrac{x-1}{x+1}+(x-1)(x+1)\dfrac{x+2}{x-1}}\\
&=\dfrac{(x+1)^2-(x-1)^2}{(x-1)^2+(x+2)(x+1)}\\
&=\dfrac{x^2+2x+1-x^2+2x-1}{x^2-2x+1+x^2+3x+2}\\
&=\dfrac{4x}{2x^2+x+3}.
\end{align*}$$
