Answer
$\dfrac{4a+1}{(3a+1)(3a-1)}$
Work Step by Step
Factoring the given expression, $
\dfrac{a}{9a^2-1}+\dfrac{2}{6a-2}
,$ results to
\begin{array}{l}\require{cancel}
\dfrac{a}{(3a+1)(3a-1)}+\dfrac{2}{2(3a-1)}
.\end{array}
Using the $LCD=
2(3a+1)(3a-1)
$, the expression above simplifies to
\begin{array}{l}
\dfrac{2(a)+(3a+1)(2)}{2(3a+1)(3a-1)}
\\\\=
\dfrac{2a+6a+2}{2(3a+1)(3a-1)}
\\\\=
\dfrac{8a+2}{2(3a+1)(3a-1)}
\\\\=
\dfrac{2(4a+1)}{2(3a+1)(3a-1)}
\\\\=
\dfrac{\cancel{2}(4a+1)}{\cancel{2}(3a+1)(3a-1)}
\\\\=
\dfrac{4a+1}{(3a+1)(3a-1)}
.\end{array}
