I have a problem with the equation which is a part of the solution: $$ y' = \frac {1 +py}{1 + p^2} $$
I tried to use $y=uv, y'=u'v + uv'$ substitutions (Bernouilli's method): $$ u'v + uv' - \frac {p}{1 + p^2} uv = \frac {1} {1+p^2}$$
By solving $v' - \frac {p}{1 +p^2}v = 0 $ I have found $v = \sqrt {1 + p^2}$. Than I have: $$u'\sqrt{1 +p^2} = \frac {1}{1+p^2} \implies u = \int \frac{dp}{\sqrt{(1+p^2)^3}} $$
Here I'm stuck but I know the solution should be: $$y = \sqrt{1+p^2}\left(\frac {p}{\sqrt{1+p^2}} + C \right) $$