The integral you're asking for is:

$\int \frac{x}{x^2 + 3} \, dx$

To solve this, we can use a substitution method.

Step-by-Step Solution:

1. Substitute: Let $u = x^2 + 3$, so $du = 2x \, dx$.

2. Rearrange: Now, divide both sides by 2 to isolate $x \, dx$: $\frac{du}{2} = x \, dx$

3. Rewrite the integral: Substituting this into the original integral, we get: $\int \frac{x}{x^2 + 3} \, dx = \int \frac{1}{2} \frac{du}{u}$

4. Solve the integral: Now, the integral becomes: $\frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln |u| + C$

5. Substitute back: Since $u = x^2 + 3$, we substitute back to get the final result: $\frac{1}{2} \ln |x^2 + 3| + C$

Thus, the integral of $\frac{x}{x^2 + 3} \, dx$ is:

$\boxed{\frac{1}{2} \ln |x^2 + 3| + C}$

Would you like further details on substitution or solving techniques?

Here are 5 related questions to deepen your understanding:

1. What is the integral of $\frac{x}{x^2 + 1}$?
2. How would you integrate $\frac{1}{x^2 + 3}$?
3. Can you generalize this method for $\frac{x}{x^2 + a}$?
4. How would the integration change if the expression was $\frac{x^3}{x^2 + 3}$?
5. How do you evaluate an improper integral involving $\frac{x}{x^2 + 3}$ over an infinite domain?

Tip: Substitution is one of the most powerful techniques in calculus, especially for rational functions. Look for expressions where the derivative of part of the function appears elsewhere in the integral.