Answer
See explanation
Work Step by Step
We take $v=v(P, T)$. Its total differential is $$
d v=\left(\frac{\partial v}{\partial T}\right)_P d T+\left(\frac{\partial v}{\partial P}\right)_T d P
$$ which, for a constant pressure process, reduces to $$
d v=\left(\frac{\partial v}{\partial T}\right)_P d T
$$ Dividing by $v$, $$
\frac{d v}{v}=\frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_P d T
$$ Using the definition of $\beta$, $$
\frac{d v}{v}=\beta d T
$$ Taking $\beta$ to be a constant, integration from 1 to 2 yields $$
\ln \frac{v_2}{v_1}=\beta\left(T_2-T_1\right)
$$ or $$
\frac{v_2}{v_1}=\exp \left[\beta\left(T_2-T_1\right)\right]
$$ which is the desired relation.
