Integral of x*e^(-3x^2) dx

1. Let $u = e^{- 3 x^{2}}$.

   Then let $du = - 6 x e^{- 3 x^{2}} dx$ and substitute $- \frac{du}{6}$:

   $$\int \frac{1}{36}\, du$$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \left(- \frac{1}{6}\right)\, du = - \frac{\int 1\, du}{6}$$

      1. The integral of a constant is the constant times the variable of integration:

         $$\int 1\, du = u$$

      So, the result is: $- \frac{u}{6}$

   Now substitute $u$ back in:

   $$- \frac{e^{- 3 x^{2}}}{6}$$

2. Add the constant of integration:

   $$- \frac{e^{- 3 x^{2}}}{6}+ \mathrm{constant}$$

The answer is:

$$- \frac{e^{- 3 x^{2}}}{6}+ \mathrm{constant}$$