Integral of e^-x(x-1) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Use integration by parts:

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

      Let $u{\left(x \right)} = x - 1$ and let $\operatorname{dv}{\left(x \right)} = e^{- x}$.

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

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

      1. There are multiple ways to do this integral.

         Method #1

         1. Let $u = - x$.

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

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

            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: $- e^{u}$

            Now substitute $u$ back in:

            $$- e^{- x}$$

         Method #2

         1. Let $u = e^{- x}$.

            Then let $du = - e^{- x} dx$ and substitute $- du$:

            $$\int 1\, du$$

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

               $$\int \left(-1\right)\, du = - \int 1\, du$$

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

                  $$\int 1\, du = u$$

               So, the result is: $- u$

            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 $- du$:

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

         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: $- e^{u}$

         Now substitute $u$ back in:

         $$- e^{- x}$$

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

   Method #2

   1. Rewrite the integrand:

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

   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. 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 $- du$:

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

            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: $- e^{u}$

            Now substitute $u$ back in:

            $$- e^{- x}$$

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

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

2. Now simplify:

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

3. Add the constant of integration:

   $$- x e^{- x}+ \mathrm{constant}$$

The answer is:

$$- x e^{- x}+ \mathrm{constant}$$