Answer
$\ln |\ln (y)|=\tan^{-1} x+\ln (2)$
Work Step by Step
Separate the differentials and integrate.
$\int (\ln y)^{-1} (\dfrac{1}{y}) dy=\int \dfrac{dx}{1+x^2}$
This implies that $\ln |\ln (y)|=\tan^{-1} x+c$
Applying the initial conditions $y(0)=e^2$, we get
Thus, $\ln |\ln (e^2)|=\tan^{-1} (0)+c\implies c=\ln (2)$
Hence, $\ln |\ln (y)|=\tan^{-1} x+\ln (2)$
