Solution of the differential equation $\displaystyle \frac{dx}{dy}-\frac{x\log x}{1+\log x}=\frac{e^{y}}{1+\log x},\ \text{ if } y(1)=0,$ is –

Treat $x$ as a function of $y$ and set $u=1+\log x\ (\Rightarrow x=e^{u-1},\ \frac{dx}{dy}=x\frac{du}{dy})$. The DE becomes $u\frac{du}{dy}=u-1+e^{\,1+y-u}.$ This transforms to a first-order non-linear equation in $u(y)$ whose implicit integral does **not** reduce to any of the listed closed forms (A)–(C). With the initial condition $y(1)=0$ (i.e., $u=1$ at $y=0$), the solution is an implicit relation not matching (A)–(C).