Ans: D
$\begin{array}{cl}
& \dfrac{1}{2x-3} + \dfrac{1}{2x+3} \\
= & \dfrac{(2x+3) + (2x-3)}{(2x-3)(2x+3)} \\
= & \dfrac{4x}{4x^2 – 9}
\end{array}$
$\begin{array}{cl}
& \dfrac{1}{2x-3} + \dfrac{1}{2x+3} \\
= & \dfrac{(2x+3) + (2x-3)}{(2x-3)(2x+3)} \\
= & \dfrac{4x}{4x^2 – 9}
\end{array}$
