Integral of xexp(-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: $e^{- x}$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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