Problem 36 - Calculus: Early Transcendentals

Question

Evaluate the indefinite integral. \( \displaystyle \int \frac{2^t}{2^t + 3} \, dt \)

Step-by-Step Solution

Answer

\( \frac{1}{\ln(2)} \ln|2^t + 3| + C \)

Step 1: Identify Substitution Variable

To evaluate the integral \( \int \frac{2^t}{2^t + 3} \, dt \), we recognize that a substitution can simplify the integrand. Let \( u = 2^t + 3 \), which makes \( 2^t = u - 3 \).

Step 2: Differentiate the Substitution Variable

Differentiate \( u \) with respect to \( t \) to find \( du \). Since \( u = 2^t + 3 \), we have \( \frac{du}{dt} = 2^t \ln(2) \). Therefore, \( du = 2^t \ln(2) \, dt \) or \( dt = \frac{du}{2^t \ln(2)} \).

Step 3: Modify the Integral with Substitution

Substitute \( u = 2^t + 3 \) and \( dt = \frac{du}{2^t \ln(2)} \) into the integral. The integral becomes:\[ \int \frac{2^t}{u} \cdot \frac{du}{2^t \ln(2)} = \int \frac{1}{u \ln(2)} \, du \]

Step 4: Simplify and Integrate

The integral \( \int \frac{1}{u \ln(2)} \, du \) simplifies to \( \frac{1}{\ln(2)} \int \frac{1}{u} \, du \). This is a standard integral: \( \int \frac{1}{u} \, du = \ln|u| + C \). Thus, we have:\[ \frac{1}{\ln(2)} \ln|u| + C \]

Step 5: Back Substitute Original Variable

Substitute back \( u = 2^t + 3 \) into the solution to express it in terms of \( t \):\[ \frac{1}{\ln(2)} \ln|2^t + 3| + C \]

Key Concepts

U-Substitution

U-substitution is a powerful tool in integration, especially when dealing with complex functions. It's a method used to simplify the integrand, which is the expression inside the integral.
Think of it as changing the variable to make integration easier. The idea is to choose a substitution for part of the integrand that complicates the integration process.
In our exercise, we used the substitution:

• Let \( u = 2^t + 3 \).
• This choice tells us that the difficult part \( 2^t + 3 \) can be replaced by \( u \), simplifying our work.

By differentiating \( u \) with respect to \( t \), you can express \( dt \) in terms of \( du \), making it easier to integrate. It's like translating a foreign language known as the integrand into one that's more familiar and simpler, allowing the integral to be evaluated with standard techniques.

Integrand Simplification

Integrand simplification involves rewriting the integrand to make it easier to integrate. When an integrand is expressed in a form that's unfriendly for integration, we use substitution to reshape it.
For example:
You start with the integral \( \int \frac{2^t}{2^t + 3} \, dt \).
By substituting \( u = 2^t + 3 \), our integrand becomes \( \int \frac{1}{u \ln(2)} \, du \).
This transformation simplifies the original expression, making it much easier to work with.
The goal is always to create an integrand that matches a familiar and easier structure you can integrate effortlessly.
Simplifying the integrand is often an essential step in providing a pathway to solve the integral systematically.

Standard Integral

Standard integrals are the backbone of calculus. They are well-established integrals that have been solved before, and their solutions are memorized because they occur frequently.
For our simplified integrand, after substitution and simplification, we arrived at the standard integral:

• \( \int \frac{1}{u} \, du \).

This integral corresponds to the natural logarithm function, leading to:

• \( \ln|u| + C \).

Where \( C \) is the constant of integration.
Using these standard integrals cuts down the work since their solutions are well-known and ready-made, which provides an efficient way to solve more complex integrals once they are simplified.

Integration Techniques

Integration techniques encompass a wide range of methods used to solve integrals, with u-substitution being a prominent one.
This technique is essential when dealing with integrals that resist direct application of standard formulas.

• Through substitution, integrals become recognizable expressions.
• Simplifying integrands allows for the direct application of standard integrals.

No single technique suits all integrals. Many integration scenarios require a combination of methods for success. Techniques like substitution deepen our understanding of integrals, helping us see them as adaptable expressions rather than rigid mathematical structures.
Understanding when and how to apply these integration techniques is crucial for mastering calculus and solving integrals efficiently.