Integral of 1-x*exp(-x) dx

1. Integrate term-by-term:

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

      $$\int \left(- x e^{- x}\right)\, dx = - \int x e^{- x}\, dx$$

      1. There are multiple ways to do this integral.

         Method #1

         1. Let $u = - x$.

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

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

            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:

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

         Method #2

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

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

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

            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:

               $$- e^{- x}$$

            Now evaluate the sub-integral.

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

            $$\int \left(- e^{- x}\right)\, dx = - \int e^{- x}\, dx$$

            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:

               $$- e^{- x}$$

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

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

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

      $$\int 1\, dx = x$$

   The result is: $x + x e^{- x} + e^{- x}$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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