To evaluate the indefinite integral
\[ \int \frac{x}{3 - x^2} , dx, \]
we can use the method of substitution. Let's set:
\[ u = 3 - x^2. \]
Then, we compute the derivative of \(u\):
\[ \frac{du}{dx} = -2x \quad \Rightarrow \quad du = -2x , dx \quad \Rightarrow \quad dx = \frac{du}{-2x}. \]
Now substituting \(u\) and \(dx\) in terms of \(u\):
\[ \int \frac{x}{3 - x^2} , dx = \int \frac{x}{u} \left(-\frac{du}{2x}\right) = -\frac{1}{2} \int \frac{1}{u} , du. \]
This simplifies to:
\[ -\frac{1}{2} \ln |u| + C. \]
Now replace \(u\) back with \(3 - x^2\):
\[ -\frac{1}{2} \ln |3 - x^2| + C. \]
Thus, the indefinite integral is:
\[ \int \frac{x}{3 - x^2} , dx = -\frac{1}{2} \ln |3 - x^2| + C. \]