Use long division or synthetic division. Below we show long division.
$$ \begin{array}{r}
2x+1\phantom{-10} \\
x-3{\overline{\smash{\big)}\,2x^2-5x-1}} \\
\underline{-(2x^2-6x)} \hphantom{-9}\\
x-1\phantom{} \\
\underline{-(x-3)} \\
\end{array} $$
$$ \text{ Remainder}=2 $$
$$\frac{2x^2-5x-1}{x-3} = \boxed{2x+1 + \frac{2}{x-3}} $$
Choose an arbitary choice to test, such as \(x=0\).
$$ \frac{2x^2-5x-1}{x-3} $$
$$ \frac{2(0)^2-5(0)-1}{0-3} $$
$$ = \frac{1}{3} $$
Testing the options,
\(2(0)+1 = \dfrac{1}{3} \)
$$ 1 \ne \frac{1}{3} $$
$$2(0)-1 = \frac{1}{3} $$
$$ -1 \ne \frac{1}{3} $$
\(2(0) + \dfrac{2}{(0)-3} = \dfrac{1}{3} \)
$$ -\frac{2}{3} \ne \frac{1}{3} $$
$$2(0)+1 + \frac{2}{(0)-3} = \frac{1}{3} $$
$$ \frac{1}{3} = \frac{1}{3} \checkmark$$
We can combine the fractions in the last two answer choices and check if they are equivalent to the initial expression. Below we show verification for the correct choice:
$$ 2x +1 + \frac{2}{x-3} $$
$$ (2x+1) \left(\frac{x-3}{x-3}\right) + \frac{2}{x-3} $$
$$ \frac{(2x+1)(x-3)}{x-3}+\frac{2}{x-3} $$
$$ \frac{2x^2+x-6x-3}{x-3}+\frac{2}{x-3} $$
$$ \frac{2x^2-5x-3+2}{x-3} $$
$$ =\frac{2x^2-5x-1}{x-3} \checkmark $$