Answer
$(\dfrac{1}{2})e^{x^2}-2 \ln (\sqrt y+2)=c$
Work Step by Step
Given: $ \dfrac{1}{x} \dfrac{dy}{dx}=ye^{x^2}+2 \sqrt y e^{x^2}$
Re-arrange the given equation and integrate as follows:.
Then $ \int x e^{x^2} dx = \int \dfrac{1}{\sqrt y(\sqrt y+2)} dy$ ....(1)
Let us take the help of $u$ substitution.
plug $u=\sqrt y \implies du =\dfrac{dy}{2 \sqrt y}$ and $x^2 =p \implies dp=2x dx$
As we know that $\int x^n dx=\dfrac{x^{n+1}}{n+1}+c$
Equation (1) becomes: $ (\dfrac{1}{2}) \int 2 x e^{x^2} dx = \int \dfrac{1}{\sqrt y(\sqrt y+2)} dy \\ (\dfrac{1}{2}) \int e^{p} dp = \int \dfrac{2 du}{u+2}$
$(\dfrac{1}{2})e^{x^2} =2 \ln (\sqrt y+2)+c$
Hence, $(\dfrac{1}{2})e^{x^2}-2 \ln (\sqrt y+2)=c$
