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

1. Use integration by parts:

   $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

   Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = e^{x^{2}}$.

   Then $\operatorname{du}{\left(x \right)} = 1$.

   To find $v{\left(x \right)}$:

   ErfRule(a=1, b=0, c=0, context=exp(x**2), symbol=x)

   Now evaluate the sub-integral.

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

   $$\int \frac{\sqrt{\pi} \operatorname{erfi}{\left(x \right)}}{2}\, dx = \frac{\sqrt{\pi} \int \operatorname{erfi}{\left(x \right)}\, dx}{2}$$

   1. Don't know the steps in finding this integral.

      But the integral is

      $$x \operatorname{erfi}{\left(x \right)} - \frac{e^{x^{2}}}{\sqrt{\pi}}$$

   So, the result is: $\frac{\sqrt{\pi} \left(x \operatorname{erfi}{\left(x \right)} - \frac{e^{x^{2}}}{\sqrt{\pi}}\right)}{2}$

3. Now simplify:

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

4. Add the constant of integration:

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

The answer is:

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