Answer
See explanation
Work Step by Step
We take $v=v(P, T)$. Its total differential is $$
d v=\left(\frac{\partial v}{\partial T}\right)_P d T+\left(\frac{\partial v}{\partial P}\right)_T d P
$$ Dividing by $v$, $$
\frac{d v}{v}=\frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_P d T+\frac{1}{v}\left(\frac{\partial v}{\partial P}\right)_T d P
$$ Using the definitions of $\alpha$ and $\beta$, $$
\frac{d v}{v}=\beta d T-\alpha d P
$$ Taking $\alpha$ and $\beta$ to be constants, integration from 1 to 2 yields $$
\ln \frac{v_2}{v_1}=\beta\left(T_2-T_1\right)-\alpha\left(P_2-P_1\right)
$$which is the desired relation.
