derivation of electric potential at a point on axial line of an electric dipole

Suppose P is a point on the axial position of the dipole.
Length of dipole=2a
Suppose point P is at the distance 'r' from the center of the dipole.
We know that potential at a point is given as,
$\mathrm{V}=\frac{1}{4\mathrm{\pi }{\in }_0}.\frac{\mathrm{Q}}{\mathrm{r}}$
So, potential at P due to q is,
${\mathrm{V}}_{\mathrm{q}}=\frac{1}{4\mathrm{\pi }{\in }_0}.\frac{\mathrm{q}}{(\mathrm{a}+\mathrm{r})}$
potential at P due to –q is,
${\mathrm{v}}_{-\mathrm{q}}=\frac{1}{4\mathrm{\pi }{\in }_0}.\frac{-\mathrm{q}}{(\mathrm{r}-\mathrm{a})}$
Total potential at P is
$$\begin{gathered} \mathrm{V}={\mathrm{V}}_{\mathrm{q}}+{\mathrm{V}}_{-\mathrm{q}} \\ =\frac{1}{4\mathrm{\pi }{\in }_0}.\frac{\mathrm{q}}{(\mathrm{a}+\mathrm{r})}+\frac{1}{4\mathrm{\pi }{\in }_0}.\frac{-\mathrm{q}}{(\mathrm{r}-\mathrm{a})} \\ =\frac{\mathrm{q}}{4\mathrm{\pi }{\in }_0}[\frac{1}{(\mathrm{a}+\mathrm{r})}+\frac{1}{(\mathrm{a}-\mathrm{r})}] \\ \mathrm{V}=\frac{\mathrm{q}}{4\mathrm{\pi }{\in }_0}.\frac{2\mathrm{a}}{({\mathrm{a}}^2-{\mathrm{r}}^2)} \end{gathered}$$