What is $\int \frac{1}{\sqrt{25y^2-10y-3}}dy$ $= \int \dfrac{1}{\sqrt{(5y-1)^2-4}}dy$
$=\int \dfrac{1}{\sqrt{u^2-4}}\dfrac{du}{5}, \quad U$ substitution 
$=\int \dfrac{1}{10\cos(\theta)} 2\cos(\theta) d\theta, \quad$ Trig substitution 
$= \dfrac{1}{5} \theta$ 
$= \dfrac{\cos^{-1}\frac{\sqrt{(5y-1)^2-4}}{2}}{5} +C$ 
Where did I go wrong?
work after u substitution $\frac{1}{5}\int \sec \theta d\theta \\ =\frac{1}{5} \ln \left |  \sec \theta + \tan \theta\right | $
* New Answer* $\\
\frac{1}{5}\ln \left | \frac{5y-1}{2} +\frac{\sqrt{(5y-1)^2 -4}}{2}\right | +C$

The problem is in the 3rd line. $$\sqrt{4\sin^2\theta-4} \neq 2\cos \theta$$
Try $x=2\cosh\theta$ or $x=2\sec\theta$
