Talk:Primitive of Reciprocal of Root of a x + b by Root of p x + q

I would tentatively suggest the problem with this proof is that $\dfrac p a$ could be negative. In that case you'd have $\ds \sqrt {-\frac p a} \sqrt {\paren {\frac {b p - a q} p} - u^2}$ hence the $\arcsin$ coming about. Haven't gone through it properly but that'd be my bet. Caliburn (talk) 11:57, 7 August 2021 (UTC)
 * (and indeed on Primitive of x squared by Root of a squared minus x squared we have $a = 1$, $b = 0$, $p = -1$, $q = a^2$, so this would seem to be the faulty case) Caliburn (talk) 12:00, 7 August 2021 (UTC)


 * Are you able to clean up this mess I made? You'd be a hero. --prime mover (talk) 13:38, 7 August 2021 (UTC)