Integral of xe^(1-x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

      $$e^{1 - x} x = e x e^{- x}$$

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

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

      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:

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

      So, the result is: $e \left(- x e^{- x} - e^{- x}\right)$

   Method #2

   1. Rewrite the integrand:

      $$e^{1 - x} x = e x e^{- x}$$

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

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

      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:

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

      So, the result is: $e \left(- x e^{- x} - e^{- x}\right)$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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