Integral of 3^(2x-1) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 2 x - 1$.

      Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

      $$\int \frac{3^{u}}{2}\, du$$

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

         $$\int 3^{u}\, du = \frac{\int 3^{u}\, du}{2}$$

         1. The integral of an exponential function is itself divided by the natural logarithm of the base.

            $$\int 3^{u}\, du = \frac{3^{u}}{\log{\left(3 \right)}}$$

         So, the result is: $\frac{3^{u}}{2 \log{\left(3 \right)}}$

      Now substitute $u$ back in:

      $$\frac{3^{2 x - 1}}{2 \log{\left(3 \right)}}$$

   Method #2

   1. Rewrite the integrand:

      $$3^{2 x - 1} = \frac{3^{2 x}}{3}$$

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

      $$\int \frac{3^{2 x}}{3}\, dx = \frac{\int 3^{2 x}\, dx}{3}$$

      1. Let $u = 2 x$.

         Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

         $$\int \frac{3^{u}}{2}\, du$$

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

            $$\int 3^{u}\, du = \frac{\int 3^{u}\, du}{2}$$

            1. The integral of an exponential function is itself divided by the natural logarithm of the base.

               $$\int 3^{u}\, du = \frac{3^{u}}{\log{\left(3 \right)}}$$

            So, the result is: $\frac{3^{u}}{2 \log{\left(3 \right)}}$

         Now substitute $u$ back in:

         $$\frac{3^{2 x}}{2 \log{\left(3 \right)}}$$

      So, the result is: $\frac{3^{2 x}}{6 \log{\left(3 \right)}}$

   Method #3

   1. Rewrite the integrand:

      $$3^{2 x - 1} = \frac{3^{2 x}}{3}$$

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

      $$\int \frac{3^{2 x}}{3}\, dx = \frac{\int 3^{2 x}\, dx}{3}$$

      1. Let $u = 2 x$.

         Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

         $$\int \frac{3^{u}}{2}\, du$$

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

            $$\int 3^{u}\, du = \frac{\int 3^{u}\, du}{2}$$

            1. The integral of an exponential function is itself divided by the natural logarithm of the base.

               $$\int 3^{u}\, du = \frac{3^{u}}{\log{\left(3 \right)}}$$

            So, the result is: $\frac{3^{u}}{2 \log{\left(3 \right)}}$

         Now substitute $u$ back in:

         $$\frac{3^{2 x}}{2 \log{\left(3 \right)}}$$

      So, the result is: $\frac{3^{2 x}}{6 \log{\left(3 \right)}}$

2. Now simplify:

   $$\frac{3^{2 x}}{6 \log{\left(3 \right)}}$$

3. Add the constant of integration:

   $$\frac{3^{2 x}}{6 \log{\left(3 \right)}}+ \mathrm{constant}$$

The answer is:

$$\frac{3^{2 x}}{6 \log{\left(3 \right)}}+ \mathrm{constant}$$