Integral of e^(3x-1)dx dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

      $$e^{3 x - 1} \cdot 1 = \frac{e^{3 x}}{e}$$

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

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

      1. There are multiple ways to do this integral.

         Method #1

         1. Let $u = 3 x$.

            Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

            $$\int \frac{e^{u}}{9}\, du$$

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

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

               1. The integral of the exponential function is itself.

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

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

            Now substitute $u$ back in:

            $$\frac{e^{3 x}}{3}$$

         Method #2

         1. Let $u = e^{3 x}$.

            Then let $du = 3 e^{3 x} dx$ and substitute $\frac{du}{3}$:

            $$\int \frac{1}{9}\, du$$

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

               $$\int \frac{1}{3}\, du = \frac{\int 1\, du}{3}$$

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

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

               So, the result is: $\frac{u}{3}$

            Now substitute $u$ back in:

            $$\frac{e^{3 x}}{3}$$

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

   Method #2

   1. Rewrite the integrand:

      $$e^{3 x - 1} \cdot 1 = \frac{e^{3 x}}{e}$$

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

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

      1. Let $u = 3 x$.

         Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

         $$\int \frac{e^{u}}{9}\, du$$

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

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

            1. The integral of the exponential function is itself.

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

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

         Now substitute $u$ back in:

         $$\frac{e^{3 x}}{3}$$

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

2. Now simplify:

   $$\frac{e^{3 x - 1}}{3}$$

3. Add the constant of integration:

   $$\frac{e^{3 x - 1}}{3}+ \mathrm{constant}$$

The answer is:

$$\frac{e^{3 x - 1}}{3}+ \mathrm{constant}$$