How to prove the following:
If $$\int \frac{adx}{(ax-2)\sqrt{(ax-1)}}=\frac{y}{\sqrt{5}}$$
then show that $\frac{1}{x}=\frac{a}{2}\left(1+sech({\frac{y}{\sqrt{5}}})\right)$
Given, at $y=0$, $x=1/a$.
Approach
I have assumed $ax-1= \cosh^2z$ then $ax-2=\sinh^2z$, $adx=2\cosh{z}\sinh{z} dz$
Then we get, $$2\int \frac{dz}{\sinh {z}}=\frac{y}{\sqrt{5}}$$
But I failed to show the desired.
Let $u=\sqrt{ax-1}$, then $x=\frac{u^2+1}{a}\,\,\,$ and $\,\,\,dx=\frac{2u}{a}du$, so \begin{align} \int\frac{adx}{(ax-2)\sqrt{ax-1}}&=\int\frac{a\left(\frac{2u}{a}\right)du}{(u^2-1)u}\\ &=\int\frac{2du}{u^2-1}\\ &=\int\left(\frac{1}{u-1}-\frac{1}{u+1}\right)du \end{align}
You can do the remaining part.