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

1. Use integration by parts:

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

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

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

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

   1. There are multiple ways to do this integral.

      Method #1

      1. Let $u = - x$.

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

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

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

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

            1. The integral of the exponential function is itself.

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

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

         Now substitute $u$ back in:

         $$- e^{- x}$$

      Method #2

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

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

         $$\int 1\, du$$

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

            $$\int \left(-1\right)\, du = - \int 1\, du$$

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

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

            So, the result is: $- u$

         Now substitute $u$ back in:

         $$- e^{- x}$$

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

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

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

   1. Let $u = - x$.

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

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

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

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

         1. The integral of the exponential function is itself.

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

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

      Now substitute $u$ back in:

      $$- e^{- x}$$

   Now evaluate the sub-integral.

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

   $$\int 2 e^{- x}\, dx = 2 \int e^{- x}\, dx$$

   1. Let $u = - x$.

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

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

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

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

         1. The integral of the exponential function is itself.

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

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

      Now substitute $u$ back in:

      $$- e^{- x}$$

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

4. Now simplify:

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

5. Add the constant of integration:

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

The answer is:

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