Integral of (3-x)*exp(x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = - x$.

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

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

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

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

         1. Let $u = - u$.

            Then let $du = - du$ 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:

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

                  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: $u e^{u} - 4 e^{u}$

            Now substitute $u$ back in:

            $$- u e^{- u} - 4 e^{- u}$$

         So, the result is: $u e^{- u} + 4 e^{- u}$

      Now substitute $u$ back in:

      $$- x e^{x} + 4 e^{x}$$

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

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

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

         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. The integral of the exponential function is itself.

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

            Now evaluate the sub-integral.

         2. The integral of the exponential function is itself.

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

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

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

         $$\int 3 e^{x}\, dx = 3 \int e^{x}\, dx$$

         1. The integral of the exponential function is itself.

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

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

      The result is: $- x e^{x} + 4 e^{x}$

   Method #3

   1. Use integration by parts:

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

      Let $u{\left(x \right)} = 3 - x$ and let $\operatorname{dv}{\left(x \right)} = e^{x}$.

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

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

      1. The integral of the exponential function is itself.

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

      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. The integral of the exponential function is itself.

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

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

2. Now simplify:

   $$\left(4 - x\right) e^{x}$$

3. Add the constant of integration:

   $$\left(4 - x\right) e^{x}+ \mathrm{constant}$$

The answer is:

$$\left(4 - x\right) e^{x}+ \mathrm{constant}$$