Problem 84 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(83-92.\) $$\int \frac{3^{x}}{3-3^{x}} d x$$

Step-by-Step Solution

Answer

The integral evaluates to \(-\frac{1}{\ln 3}\ln |1 - 3^{x-1}| + C\).

Step 1: Simplify the Integral

To solve the integral \( \int \frac{3^{x}}{3-3^{x}} \, dx \), we begin by rewriting the expression in a simpler form. Notice that \( 3 - 3^x = 3(1 - 3^{x-1}) \). Consequently, the integral becomes \( \int \frac{3^{x}}{3(1-3^{x-1})} \, dx \), or \( \int \frac{3^{x-1}}{1-3^{x-1}} \, dx \). Let \( u = 3^{x-1} \), making \( du = 3^{x-1} \ln 3 \, dx \).

Step 2: Integrate using Substitution

With the substitution \( u = 3^{x-1} \), the integral becomes \( \int \frac{u}{1-u} \frac{du}{u \ln 3} = \frac{1}{\ln 3} \int \frac{1}{1-u} \, du \). This simplifies to \( -\frac{1}{\ln 3} \ln |1-u| + C \), where \( C \) is the constant of integration.

Step 3: Back-substitute to the Original Variable

Since \( u = 3^{x-1} \), substitute back to find the expression in terms of \( x \): \( -\frac{1}{\ln 3} \ln |1 - 3^{x-1}| + C \). Thus, the integral evaluates to \(-\frac{1}{\ln 3}\ln |1 - 3^{x-1}| + C\).

Key Concepts

Substitution Method

The substitution method is a powerful technique used to simplify complex integrals. It involves replacing parts of the integral with a single variable to make it easier to handle. In our exercise, the integral \( \int \frac{3^{x}}{3-3^{x}} \, dx \) contains an expression that can be simplified by substitution. We choose \( u = 3^{x-1} \) because it transforms the variable and reduces the expression to a simpler form. This step is crucial as it helps to change the variable from \( x \) to \( u \), making the integration process more straightforward. Once we've done the substitution, we also need the differential, \( du \), to replace \( dx \). With \( u = 3^{x-1} \) and \( du = 3^{x-1} \ln 3 \, dx \), we substitute and transform the integral into a simpler version, \( \int \frac{1}{1-u} \, du \), which is easier to evaluate.

Logarithmic Integration

Logarithmic integration is a special method used to solve integrals that result in a logarithmic expression. In our solution, after substitution, the integral becomes \( \int \frac{1}{1-u} \, du \), which is a standard form leading to a logarithmic result. The antiderivative of \( \frac{1}{1-u} \) is \(-\ln |1-u|\), representing the integral of a rational function. This step uses the property that \( \int \frac{1}{v} \, dv = \ln |v| \). By handling the integral in this way, we achieve an expression of \( -\frac{1}{\ln 3} \ln |1-u| + C \), with \( C \) as the constant of integration. This approach is vital as it allows us to handle the integration of functions that naturally develop into logarithmic forms.

Exponential Functions

Exponential functions involve expressions where the variable is in the exponent, such as \( 3^x \). They are encountered frequently in integrals, requiring special techniques like substitution. In our exercise, the expression \( 3^x \) is a key part of the integral challenge. When we initially rewrite \( 3-3^x \) as \( 3(1-3^{x-1}) \), it highlights how exponential functions can be manipulated algebraically to facilitate substitution. Exponential expressions possess unique properties that are useful in integration, such as scaling and transforming the interval through substitution. Understanding these principles aids in solving integrals involving exponential terms, offering a clearer path to finding an antiderivative. Through substitution, we convert the exponential function into something manageable and then solve the integral using known integration techniques.