Problem 21 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Compute the following integrals. $$ \int \frac{3^{x}}{3^{x}+1} d x $$

Step-by-Step Solution

Answer

The integral of \( \frac{3^{x}}{3^{x}+1} dx \) is \( \frac{1}{\ln(3)} \ln|3^x + 1| + C \).

Step 1: Substitute variables

Let \( u = 3^x + 1 \). Then, the derivative du/dx will be \( \ln(3) \cdot 3^x \). Moving dx to the other side, we have \( du = \ln(3) \cdot 3^x dx \). Rearranging terms, \( dx = \frac{du}{\ln(3) \cdot 3^x} \). Substitute the given equation with these new terms.

Step 2: Substitute variables into equation

Substitute dx into the original integral. We have: \[ \int \frac{3^x}{u} \cdot \frac{du}{\ln(3) \cdot 3^x} \]. The term \( 3^x \) cancels out, leaving us with: \[ \frac{1}{\ln(3)} \int \frac{1}{u} du \].

Step 3: Integrate

The integral of \( \frac{1}{u} \) is \( \ln|u| \), so we have \( \frac{1}{\ln(3)} \ln|u| + C\).

Step 4: Substitute back

Substitute back the original variables. Replace \( u \) with \( 3^x + 1 \). The final result is \( \frac{1}{\ln(3)} \ln|3^x + 1| + C \).

Key Concepts

Substitution Method

The substitution method is a clever and often-used technique for finding integrals in calculus. It involves changing the variable in the integral to simplify the integration process. In essence, you replace complex parts of the expression with a single variable to make the integration more manageable. This method is particularly useful when you encounter an integral in which the integrand includes linked parts, like a function and its derivative.In the given exercise, we started by recognizing that the denominator, \(3^x + 1\), can be replaced by a simpler variable \(u\). This changes the problem into an easier form by eliminating the complexity of the exponential function. To use the substitution method effectively:

• Identify a part of the integrand that makes solving difficult, usually involving compound expressions.
• Choose a substitution \( u = \text{expression} \) that simplifies the problem.
• Find \( du \), the differential of your substitution.
• Express \( dx \) in terms of \( du \).

The substitution method transforms the integral into a more familiar form, making the integration process easier.

Indefinite Integral

An indefinite integral represents a varied family of functions, all of which are the antiderivative of a given function. It includes a constant of integration, usually denoted as \( C \), because derivatives of constants are zero and hence do not affect the differentiation of our antiderivative solution.In our exercise, once we substituted variables and simplified the integral, we ended up needing to find the indefinite integral of \( \frac{1}{u} \), a well-known form. The indefinite integral of \( \frac{1}{u} \) is \( \ln|u| \). So, after integration, we appended the constant \( C \) to our result:

• Indefinite integrals can be interpreted geometrically as the family of all antiderivatives of a function.
• The constant \( C \) represents any constant number, because differentiating a constant results in zero.
• This makes the indefinite integral suitable for problems requiring a general solution.

Thus, the indefinite integral allows us to backtrack to the original function from its derivative, with room for a constant added.

U-Substitution

U-substitution is a specific application of the substitution method that introduces a new variable \( u \) to simplify the integration. In practical terms, it's like saying "this piece of the problem is too complicated to work with directly, so let's rewrite the problem using \( u \)."In this exercise, we use \( u = 3^x + 1 \). By choosing \( u \) wisely, the integral's structure becomes more straightforward. The derivative \( du = \ln(3) \cdot 3^x \/ dx \) allows us to substitute \( dx \) effectively, streamlining the calculation.Here's how u-substitution simplifies our integral:

• Substitute \( u = 3^x + 1 \) and find the corresponding \( du \).
• Rewrite the integral, replacing all occurrences of \( x \) with \( u \).
• The original messy expression becomes an integral of \( \frac{1}{u} \).

After simplifying, we found the integral solution and substituted back to the original variable, producing a more digestible end result. U-substitution hence transforms challenging integrals into simpler, more recognizable forms.