Answer

Substitute \(v = \diffone{y}\). Rearranging,

\[ \begin{aligned} \diffone{v} &= 1 + v^{2}\newline \frac{\diffone{v}}{1 + v^{2}} &= 1\newline \implies \tan^{-1} v &= x + c\newline v &= tan \roundbr{x + c}\newline \diffone{y} &= tan \roundbr{x + c}\newline \implies y &= -\ln \roundbr{\cos \roundbr{x + c}} \end{aligned} \]