Answer
$\dfrac{3}{2}x^3+\dfrac{1}{4}x^2+\dfrac{1}{8}x-\dfrac{7}{16}+\dfrac{1}{16(2x-1)}$
Work Step by Step
Using the long division below, then
\begin{array}{l}\require{cancel}\left(
3x^4-x-x^3+\dfrac{1}{2}
\right)\div(
2x-1
)\\\\=
\dfrac{3}{2}x^3+\dfrac{1}{4}x^2+\dfrac{1}{8}x-\dfrac{7}{16}+\dfrac{1}{16(2x-1)}
.\end{array}