1. Given differential equation

   $\frac{dy}{dx} - \frac{y+3x}{\ln(y+3x)} + 3 = 0$

   Rearranging,

   $\frac{dy}{dx}+3 = \frac{y+3x}{\ln(y+3x)}$

2. Use substitution

   Let

   $v = y+3x$

   Then,

   $\frac{dv}{dx} = \frac{dy}{dx}+3$

   So the differential equation becomes

   $\frac{dv}{dx} = \frac{v}{\ln v}$

3. Separate variables

   $\frac{\ln v}{v}\,dv = dx$

4. Integrate both sides

   $\int \frac{\ln v}{v}\,dv = \int dx$

   Using the standard result,

   $\int \frac{\ln v}{v}\,dv = \frac{1}{2}(\ln v)^2$

   Hence,

   $\frac{1}{2}(\ln v)^2 = x + C$

   Rearranging,

   $x - \frac{1}{2}(\ln v)^2 = C$

5. Substitute back

   Since $v=y+3x$,

   $x - \frac{1}{2}\left(\ln(y+3x)\right)^2 = C$

6. Match with options

   This is exactly Option A.

Verification with stored answer: Stored correct answer is A, which matches our derived answer.