To solve the integral $\int_0^\infty x e^{-6x^2} \, dx$, we can use a standard method for integrals of this type, which involves recognizing it as a Gaussian integral, but with a modification due to the $x$ term in the integrand.

Steps:

1. Substitution: Let $u = 6x^2$, so that: $du = 12x \, dx \quad \text{or} \quad x \, dx = \frac{du}{12}.$

2. Rewriting the Integral: The limits change as follows:

   • When $x = 0$, $u = 0$.
   • When $x = \infty$, $u = \infty$.

   Now, the integral becomes: $\int_0^\infty x e^{-6x^2} \, dx = \int_0^\infty e^{-u} \frac{du}{12}.$

3. Simplifying: $\frac{1}{12} \int_0^\infty e^{-u} \, du.$

4. Solving the Integral: The integral $\int_0^\infty e^{-u} \, du$ is a standard exponential integral, and its value is 1.

   Therefore: $\frac{1}{12} \times 1 = \frac{1}{12}.$

Final Answer:

$\int_0^\infty x e^{-6x^2} \, dx = \frac{1}{12}.$

Would you like a more detailed explanation or help with another problem? Here are some related questions you might find interesting:

1. What is the general method to solve integrals involving Gaussian functions?
2. How do we apply substitution to integrals of this form?
3. Can this method be generalized to integrals with different powers of $x$?
4. What is the relationship between Gaussian integrals and probability theory?
5. How do you handle improper integrals like this one, with infinite limits?

Tip: When dealing with integrals of the form $\int_0^\infty x e^{-ax^2} \, dx$, always look for substitution methods or recognize it as a Gaussian integral with an additional linear term.