The solution of the differential equation $$\frac{dy}{dx} - \frac{y+3x}{\log_e(y+3x)} + 3 = 0$$ is (where C is a constant of integration)

We begin with the given differential equation

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

First, we separate the derivative from the remaining terms:

$$\frac{dy}{dx}= \frac{y+3x}{\log_e(y+3x)}-3.$$

At this point it is very convenient to introduce a new variable that appears naturally in the fraction. We set

$$t = y+3x.$$

Differentiating both sides with respect to $$x$$ gives

$$\frac{dt}{dx}= \frac{d}{dx}(y+3x)=\frac{dy}{dx}+3.$$

Now we already have an expression for $$\dfrac{dy}{dx}$$ from the differential equation, so we substitute it:

$$\frac{dt}{dx}= \left(\frac{y+3x}{\log_e(y+3x)}-3\right)+3.$$

Simplifying the right side, the two “$$-3$$” terms cancel:

$$\frac{dt}{dx}= \frac{y+3x}{\log_e(y+3x)}.$$

But $$y+3x$$ is exactly our new variable $$t$$. Hence we have a much simpler differential equation,

$$\frac{dt}{dx}= \frac{t}{\log_e t}.$$

This form is separable. We rewrite it by bringing all $$t$$-terms to the left and the $$x$$-term to the right:

$$\frac{\log_e t}{t}\,dt = dx.$$

Next, we integrate both sides. For the left‐hand side we will make an explicit substitution. Let

$$u = \log_e t \quad\Longrightarrow\quad t = e^u,\quad dt = e^u\,du.$$

Then

$$\frac{\log_e t}{t}\,dt \;=\; \frac{u}{e^u}\cdot e^u\,du \;=\; u\,du.$$

Now we use the standard integral formula

$$\int u\,du = \frac{u^2}{2} + \text{constant}.$$

Therefore, integrating both sides gives

$$\int \frac{\log_e t}{t}\,dt = \int dx \quad\Longrightarrow\quad \frac{u^2}{2} = x + C_1,$$

where $$C_1$$ is a constant of integration. Substituting back $$u = \log_e t$$ we obtain

$$\frac{(\log_e t)^2}{2} = x + C_1.$$

Now we restore the original variable $$t = y+3x$$:

$$\frac{(\log_e(y+3x))^2}{2} = x + C_1.$$

It is customary to collect the constant on one side. Multiplying by $$1$$ does not change the expression, so we write

$$x - \frac{1}{2}\bigl(\log_e(y+3x)\bigr)^2 = -C_1.$$

Since $$-C_1$$ is just another arbitrary constant, we rename it as $$C$$. Thus the implicit solution finally reads

$$x - \frac{1}{2}\bigl(\log_e(y + 3x)\bigr)^2 = C.$$

This matches exactly Option A.

Hence, the correct answer is Option A.