#### Answer

$\dfrac{x^2-3x+10}{2x^2-18}$

#### Work Step by Step

Factoring the given expression, $ \dfrac{x}{2x+6}+\dfrac{5}{x^2-9} ,$ results to \begin{array}{l}\require{cancel} \dfrac{x}{2(x+3)}+\dfrac{5}{(x+3)(x-3)} .\end{array} Using the $LCD= 2(x+3)(x-3) $, the expression above simplifies to \begin{array}{l} \dfrac{(x-3)(x)+2(5)}{2(x+3)(x-3)} \\\\= \dfrac{x^2-3x+10}{2(x+3)(x-3)}
\\\\=
\dfrac{x^2-3x+10}{2(x^2-9)}
\\\\=
\dfrac{x^2-3x+10}{2x^2-18}
.\end{array}