\[\begin{align*} \boxed{\int 2^x \, dx = \frac{2^x}{\ln(2)} + C} \end{align*}\]
where \( C \) is the constant of integration.
Step-by-Step Solution:
\[\begin{align*} \int 2^x \, dx \end{align*}\]
Since \( 2^x \) is an exponential function with a base other than \( e \), to simplify the integration process, we can rewrite \( 2^x \) in terms of \( e \) using the identity:
\[\begin{align*} 2^x = e^{x \ln(2)} \end{align*}\]
This is valid because \( e^{x \ln(2)} \) is equivalent to \( 2^x \), as \( e^{\ln(2)} = 2 \). substituting \( e^{x \ln(2)} \) into the integral:
\[\begin{align*} \int 2^x \, dx = \int e^{x \ln(2)} \, dx \end{align*}\]
To integrate \( e^{x \ln(2)} \), we use the formula for the integral of an exponential function \( e^{ax} \), which is \( \frac{e^{ax}}{a} \). Here, \( a = \ln(2) \), so we get:
\[\begin{align*} \int e^{x \ln(2)} \, dx = \frac{e^{x \ln(2)}}{\ln(2)} + C \end{align*}\]
Replacing \( e^{x \ln(2)} \) with \( 2^x \), we get that the integral of \( 2^x \) is:
\[\begin{align*} \boxed{\int 2^x \, dx = \frac{2^x}{\ln(2)} + C} \end{align*}\]
Why This Works:
The reason \( \ln(2) \) appears in the denominator is that it accounts for the rate of growth of \( 2^x \) compared to \( e^x \). Integrating exponential functions with a base other than \( e \) requires dividing by the natural logarithm of the base to adjust for this rate difference.
