Integral of -x*exp(-x) dx

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. 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:

   $$- e^{- x}$$

3. Now simplify:

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

4. Add the constant of integration:

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

The answer is:

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