Answer
See the detailed answer below.
Work Step by Step
We can divide the hollow cylinder shell into very thin rings, as we see in the figure below.
The electric potential of one ring [the black ring] of thickness $dy$ and a net charge of $dy$ at the center of the cylinder is given by
$$dV_{\rm ring}=\dfrac{1}{4\pi \epsilon_0}\dfrac{dq}{\sqrt{R^2+y^2}}\tag1$$
Assuming that the charge density of the hollow cylinder is $\eta$, so
$$\eta=\dfrac{Q}{2\pi R L}=\dfrac{dq}{dA}$$
where $dA$ of a ring is given by $2\pi r dy$
$$\eta=\dfrac{Q}{2\pi R L}=\dfrac{dq}{2\pi R dy}$$
Thus,
$$dq=\dfrac{Qdy}{L}$$
Plug into (1),
$$dV_{\rm ring}=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q}{L\sqrt{R^2+y^2}} dy$$
$$dV_{\rm ring}=\dfrac{Q}{(4\pi \epsilon_0)L}\dfrac{dy}{ \sqrt{R^2+y^2}} $$
The net electric potential of the whole cylinder is then given by
$$ V_{\rm cylinder}=\int dV_{\rm ring}=\int_{-L/2}^{L/2}\dfrac{Q}{(4\pi \epsilon_0)L}\dfrac{dy}{ \sqrt{R^2+y^2}} $$
$$ V_{\rm cylinder} =\dfrac{2Q}{(4\pi \epsilon_0)L}\int_{0}^{L/2}\dfrac{dy}{ \sqrt{R^2+y^2}} $$
$$ V_{\rm cylinder} =\dfrac{2Q}{(4\pi \epsilon_0)L}\ln\left(y+\sqrt{R^2+y^2}\right)_{0}^{L/2} $$
$$ V_{\rm cylinder} =\dfrac{2Q}{(4\pi \epsilon_0)L}\ln\left(\dfrac{\dfrac{L}{2}+\sqrt{R^2+\dfrac{L^2}{4}}}{R}\right) $$
$$\boxed{ V_{\rm cylinder} =\dfrac{2Q}{(4\pi \epsilon_0)L}\ln\left( \dfrac{L}{2R}+\sqrt{1+\dfrac{L^2}{4R^2}} \right) }$$
