Integral of (x-3)e^-x dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = - x$.

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

      $$\int \left(u e^{u} + 3 e^{u}\right)\, du$$

      1. Integrate term-by-term:

         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.

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

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

            $$\int 3 e^{u}\, du = 3 \int e^{u}\, du$$

            1. The integral of the exponential function is itself.

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

            So, the result is: $3 e^{u}$

         The result is: $u e^{u} + 2 e^{u}$

      Now substitute $u$ back in:

      $$- x e^{- x} + 2 e^{- x}$$

   Method #2

   1. Rewrite the integrand:

      $$e^{- x} \left(x - 3\right) = x e^{- x} - 3 e^{- x}$$

   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:

         $$\int \left(- 3 e^{- x}\right)\, dx = - 3 \int e^{- x}\, dx$$

         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: $3 e^{- x}$

      The result is: $- x e^{- x} + 2 e^{- x}$

2. Now simplify:

   $$\left(2 - x\right) e^{- x}$$

3. Add the constant of integration:

   $$\left(2 - x\right) e^{- x}+ \mathrm{constant}$$

The answer is:

$$\left(2 - x\right) e^{- x}+ \mathrm{constant}$$