Integral of (x+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 $du$:

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

      1. Integrate term-by-term:

         1. Use integration by parts:

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

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

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

            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. The integral of the exponential function is itself.

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

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

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

      1. Let $u = - x$.

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

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

         1. Use integration by parts:

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

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

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

            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. The integral of the exponential function is itself.

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

         Now substitute $u$ back in:

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

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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