Let $E$ be an elliptic curve given by $$ y^2=x^3-fx^2+\frac{1}{b}x $$ and let one the cycles of the torus be $A$. Then the period integral is \begin{equation}a= \oint_A \lambda \end{equation} and this is given by the solution of $$ \frac{d \lambda}{df}=\frac{dx}{y}.$$ So, how exactly do we find $a$?
I am trying to directly put $y = \sqrt{x^3-fx^2+\frac{1}{b}x}$ in the differential equation but I do not get somewhere. I am sure there is some kind of identity helping with this. It seems a standard thing to do in the theory of elliptic curves but nowhere is done explicitly.