Integral of xexp(-x^2) dx

1. There are multiple ways to do this integral.

   Method #1

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

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

      $$\int \frac{1}{4}\, du$$

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

         $$\int \left(- \frac{1}{2}\right)\, du = - \frac{\int 1\, du}{2}$$

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

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

         So, the result is: $- \frac{u}{2}$

      Now substitute $u$ back in:

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

   Method #2

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

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

      $$\int \frac{e^{u}}{4}\, du$$

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

         $$\int \left(- \frac{e^{u}}{2}\right)\, du = - \frac{\int e^{u}\, du}{2}$$

         1. The integral of the exponential function is itself.

            $$\int e^{u}\, du = e^{u}$$

         So, the result is: $- \frac{e^{u}}{2}$

      Now substitute $u$ back in:

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

2. Add the constant of integration:

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

The answer is: