<s>[INST] Computing $\int \frac{1}{\sqrt{25y^2-30y-7}} \,dy$ $$\int \frac{1}{\sqrt{25y^2-30y-7}} \, dy$$
I start by completing the square: so that $(5y-3)^2-16=25y^2-30y-7$.
Then I let $u=5y-3$, and then let $ u=4 \sec \theta$.
This leads to the equation becoming $\frac{1}{4\tan\theta}$ which is the same as $\frac{1}{4}\cot \theta$.
Taking the integral of this is $\frac{1}{4}\ln\left|\sin\theta\right|$.
Substituting back I get $\frac{1}{4}\ln\left| \sin \left( \operatorname{arcsec} \frac{5y-3} 4 \right)\right|$.
I drew a triangle and came up with this: $\frac{1}{4}\ln\left|\frac{\sqrt{25y^2-30y-7}}{5y-3}\right|$. Please, where did I go wrong?
 [/INST] I realized my mistake! I forgot to put in the du in my substitutions. I ended up with $\frac{1}{5}\int sec \theta$ which ends up being $\frac{1}{5}ln|sec\theta+tan \theta|$ which is $\frac{1}{5}ln|\frac{5y-3+\sqrt(25y^2-30y-7)}{4}$|+C. Hooray!
</s>