91Ó°ÊÓ

The substitution method is a powerful technique in calculus used to simplify the process of integration. Essentially, it involves making a substitution to transform the integral into a simpler form. This method is particularly helpful when dealing with complex expressions or functions nested within each other.

In the given example, we began by identifying a part of the expression that made integration challenging. We chose the substitution:

• Let \( u = 1 - 3^{x-1} \)

The goal of this substitution is to simplify the expressions and eliminate the cumbersome parts. By doing so, we can express the original integral in terms of \( u \), allowing us to integrate more easily.

After substituting, we also differentiate to find \( du \):

• \( du = -\ln(3) \cdot 3^{x-1} \cdot dx \)

We solved for \( dx \) to replace it in the integral. This enables us to rewrite the original integral in terms of \( u \), ultimately simplifying the integration process.

Indefinite integrals represent a family of functions, defined without specific limits of integration. The result of an indefinite integral includes a constant of integration, usually denoted as \( C \). This constant accounts for the vertical shift of the function rather than a singular solution.

In this example, we calculated the indefinite integral:

• \( \int \frac{3^x}{3 - 3^x} dx \)

The aim here is to find a generic function (or several functions) that satisfies the condition when differentiated. Using substitution and rearranging terms, we transformed the expression into more manageable parts that could be integrated easily:

Once the integration is performed using these techniques, the constant \( C \) is included with the resulting function. The output is an expression that represents all possible antiderivatives of the integral we began with.

Logarithmic integration is a specific integration technique that comes into play when dealing with functions that involve logarithms. When integrating functions of the form \( \int \frac{1}{u} du \), the result is \( \ln|u| + C \). This reflects the inverse relationship that exists between differentiation and integration with natural logarithms.

In this particular solution, logarithmic integration was a key step. We derived the expression:

• \( \int \frac{1-u}{u} du = \int \frac{1}{u} du - \int 1 du \)

Which simplified to:

• \( \ln|u| - u + C \)

This represents the logarithmic integration part, highlighting how understanding the properties of logarithms can significantly ease the integration process.

Finally, by substituting back the expression for \( u \) in terms of \( x \), we expressed our final answer in the same variable as the original question, ensuring the context was maintained.