Integral of d(2^(2x-1)) dx

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

   $$\int 2^{2 x - 1} d\, dx = d \int 2^{2 x - 1}\, dx$$

   1. There are multiple ways to do this integral.

      Method #1

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

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

         $$\int \frac{2^{u}}{4}\, du$$

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

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

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

               $$\int 2^{u}\, du = \frac{2^{u}}{\log{\left(2 \right)}}$$

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

         Now substitute $u$ back in:

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

      Method #2

      1. Rewrite the integrand:

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

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

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

         1. Let $u = 2 x$.

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

            $$\int \frac{2^{u}}{4}\, du$$

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

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

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

                  $$\int 2^{u}\, du = \frac{2^{u}}{\log{\left(2 \right)}}$$

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

            Now substitute $u$ back in:

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

         So, the result is: $\frac{2^{2 x}}{4 \log{\left(2 \right)}}$

      Method #3

      1. Rewrite the integrand:

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

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

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

         1. Let $u = 2 x$.

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

            $$\int \frac{2^{u}}{4}\, du$$

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

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

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

                  $$\int 2^{u}\, du = \frac{2^{u}}{\log{\left(2 \right)}}$$

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

            Now substitute $u$ back in:

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

         So, the result is: $\frac{2^{2 x}}{4 \log{\left(2 \right)}}$

   So, the result is: $\frac{2^{2 x - 1} d}{2 \log{\left(2 \right)}}$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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