Relationship between solubility and enthalpy of solution
For equilibria such as $\ce{[C6H5COOH]_{(s)} &lt;=&gt; [C6H5COOH]_{(aq)} }$.  The equilibrium constant $K_{eq}$ can be written in the very simple form:

$$K_{eq}\equiv \ce{[C6H5COOH]_{(aq)}} \equiv \text{solubility}\,  (sol)$$

That's more than likely where the equation comes from. Then you get 

$$\dfrac{d \ln Sol}{dT}=\dfrac{\Delta H}{RT^2}$$ 

when you separate the variables and integrate with respect to both sides you get:

$$\int_{\ln sol_1}^{\ln sol_2}d(\ln sol) = \int_{T_1}^{T_{2}}\dfrac{\Delta H}{RT^2}\, dT$$

$$\ln sol_2 - \ln sol_1 = \dfrac{\Delta H}{R}\Bigg{(}\dfrac{1}{T_1}- \dfrac{1}{T_2}\Bigg{)}$$

Which gives the useful result

$$\ln \dfrac{sol_2}{sol_1}= \dfrac{\Delta H}{R}\Bigg{(}\dfrac{1}{T_1}- \dfrac{1}{T_2}\Bigg{)}$$