Integral of e^(5-4x) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

      $$e^{5 - 4 x} = e^{5} e^{- 4 x}$$

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

      $$\int e^{5} e^{- 4 x}\, dx = e^{5} \int e^{- 4 x}\, dx$$

      1. There are multiple ways to do this integral.

         Method #1

         1. Let $u = - 4 x$.

            Then let $du = - 4 dx$ and substitute $- \frac{du}{4}$:

            $$\int \frac{e^{u}}{16}\, du$$

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

               $$\int \left(- \frac{e^{u}}{4}\right)\, du = - \frac{\int e^{u}\, du}{4}$$

               1. The integral of the exponential function is itself.

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

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

            Now substitute $u$ back in:

            $$- \frac{e^{- 4 x}}{4}$$

         Method #2

         1. Let $u = e^{- 4 x}$.

            Then let $du = - 4 e^{- 4 x} dx$ and substitute $- \frac{du}{4}$:

            $$\int \frac{1}{16}\, du$$

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

               $$\int \left(- \frac{1}{4}\right)\, du = - \frac{\int 1\, du}{4}$$

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

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

               So, the result is: $- \frac{u}{4}$

            Now substitute $u$ back in:

            $$- \frac{e^{- 4 x}}{4}$$

      So, the result is: $- \frac{e^{5} e^{- 4 x}}{4}$

   Method #2

   1. Rewrite the integrand:

      $$e^{5 - 4 x} = e^{5} e^{- 4 x}$$

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

      $$\int e^{5} e^{- 4 x}\, dx = e^{5} \int e^{- 4 x}\, dx$$

      1. Let $u = - 4 x$.

         Then let $du = - 4 dx$ and substitute $- \frac{du}{4}$:

         $$\int \frac{e^{u}}{16}\, du$$

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

            $$\int \left(- \frac{e^{u}}{4}\right)\, du = - \frac{\int e^{u}\, du}{4}$$

            1. The integral of the exponential function is itself.

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

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

         Now substitute $u$ back in:

         $$- \frac{e^{- 4 x}}{4}$$

      So, the result is: $- \frac{e^{5} e^{- 4 x}}{4}$

2. Now simplify:

   $$- \frac{e^{5 - 4 x}}{4}$$

3. Add the constant of integration:

   $$- \frac{e^{5 - 4 x}}{4}+ \mathrm{constant}$$

The answer is:

$$- \frac{e^{5 - 4 x}}{4}+ \mathrm{constant}$$