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

1. Let $u = - x$.

   Then let $du = - dx$ and substitute $- du$:

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

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

      $$\int u^{2} e^{u}\, du = - \int u^{2} e^{u}\, du$$

      1. Use integration by parts:

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

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

         Then $\operatorname{du}{\left(u \right)} = 2 u$.

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

         1. The integral of the exponential function is itself.

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

         Now evaluate the sub-integral.

      2. Use integration by parts:

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

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

         Then $\operatorname{du}{\left(u \right)} = 2$.

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

         1. The integral of the exponential function is itself.

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

         Now evaluate the sub-integral.

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

         $$\int 2 e^{u}\, du = 2 \int e^{u}\, du$$

         1. The integral of the exponential function is itself.

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

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

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

   Now substitute $u$ back in:

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

2. Now simplify:

   $$- \left(x^{2} + 2 x + 2\right) e^{- x}$$

3. Add the constant of integration:

   $$- \left(x^{2} + 2 x + 2\right) e^{- x}+ \mathrm{constant}$$

The answer is:

$$- \left(x^{2} + 2 x + 2\right) e^{- x}+ \mathrm{constant}$$