Answer
$2 \ln (\sqrt y+1)=\ln |x|+\ln (4)$
Work Step by Step
Separate the differentials and integrate.
$\int \dfrac{1}{\sqrt y(\sqrt y+1)} dy=\int \dfrac{dx}{x}$
Re-write as: $\int \dfrac{2 dy}{2 \sqrt y(\sqrt y+1)} =\int \dfrac{dx}{x}$
This implies that $2 \ln |(\sqrt y+1)|=\ln |x|+c$
Applying the initial conditions $y(1)=1$, we get
Thus, $2 \ln |(\sqrt (1)+1)|=\ln |1|+c\implies c=2 \ln (2)=\ln (4)$
Hence, $2 \ln (\sqrt y+1)=\ln |x|+\ln (4)$
