Integral of (x^2+1)e^-x dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = - x$.

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

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

      1. Integrate term-by-term:

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

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

            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: $- u^{2} e^{u} + 2 u e^{u} - 2 e^{u}$

         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: $- e^{u}$

         The result is: $- u^{2} e^{u} + 2 u e^{u} - 3 e^{u}$

      Now substitute $u$ back in:

      $$- x^{2} e^{- x} - 2 x e^{- x} - 3 e^{- x}$$

   Method #2

   1. Rewrite the integrand:

      $$e^{- x} \left(x^{2} + 1\right) = x^{2} e^{- x} + e^{- x}$$

   2. Integrate term-by-term:

      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:

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

      1. Let $u = - x$.

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

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

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

            $$\text{False}$$

            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:

         $$- e^{- x}$$

      The result is: $- x^{2} e^{- x} - 2 x e^{- x} - 3 e^{- x}$

2. Now simplify:

   $$- \left(x^{2} + 2 x + 3\right) e^{- x}$$

3. Add the constant of integration:

   $$- \left(x^{2} + 2 x + 3\right) e^{- x}+ \mathrm{constant}$$

The answer is:

$$- \left(x^{2} + 2 x + 3\right) e^{- x}+ \mathrm{constant}$$