Problem 30 - Calculus and its applications

Question

Evaluate. (Be sure to check by differentiating!) $$ \int \frac{e^{t} d t}{3+e^{t}} $$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln |3 + e^{t}| + C \).

Step 1: Identify the integration technique

The given integral is \( \int \frac{e^{t} d t}{3+e^{t}} \). This form suggests using the substitution method because the derivative of \( e^{t} \) simplifies with the integral.

Step 2: Substitution

Let \( u = 3 + e^{t} \). Then, the derivative \( du = e^{t} dt \). Replace \( e^{t} dt \) in the integral with \( du \), changing the variable: \[ \int \frac{e^{t} dt}{3+e^{t}} = \int \frac{du}{u} \].

Step 3: Integrate

The integral \( \int \frac{du}{u} \) is a standard natural logarithm integral, resulting in \( \ln |u| + C \). Therefore, we have \( \ln |3+e^{t}| + C \) after substituting back \( u = 3 + e^{t} \).

Step 4: Differentiate to verify

To verify, differentiate \( \ln |3 + e^{t}| + C \). The derivative using the chain rule is \( \frac{1}{3 + e^{t}} \cdot e^{t} = \frac{e^{t}}{3 + e^{t}} \), which matches the original integrand. This confirms our solution.

Key Concepts

Substitution Method

The substitution method is a powerful technique used in calculus integration to simplify complex integrals by transforming them into a more manageable form. This technique essentially involves substituting a part of the integral with a new variable, which makes the integration process easier.

• First, identify a substitution that can simplify the integral. The goal is to choose a part of the integral, often the inner function of a composite function, as a new variable.
• Determine the derivative of this new variable, which will replace part of the differential in the integral.
• Rewrite the integral entirely in terms of the new variable. This usually simplifies the integral into a basic form that is easier to evaluate.
• Finally, once the integral is evaluated, substitute back the original variable to express the solution in terms of the original variable.

In the context of our exercise, we substituted by letting \( u = 3 + e^{t} \), and consequently, \( du = e^{t} dt \). This clever choice transformed the integral into \( \int \frac{du}{u} \), which is much simpler to solve.

Definite and Indefinite Integrals

Integrals are the building blocks of calculus, consisting of definite and indefinite integrals. Understanding the difference between these two types is essential.

• Indefinite Integrals: These represent a family of functions and are denoted without limits. When you solve an indefinite integral, you include an arbitrary constant \( C \), because integration is the reverse of differentiation, and differentiating a constant yields zero.
• Definite Integrals: These have limits of integration and result in a specific value. They are used to calculate the net area under a curve between two points. Unlike indefinite integrals, definite integrals do not include the constant \( C \).

For our exercise, solving \( \int \frac{e^{t} dt}{3+e^{t}} \) corresponds to finding an indefinite integral since no limits are given. Hence, the solution is presented as \( \ln |3+e^{t}| + C \), where \( C \) is an arbitrary constant.

Natural Logarithm Integration

Natural logarithmic functions are integral components of calculus, especially when dealing with integration of rational functions. The integral \( \int \frac{1}{u} \, du = \ln |u| + C \) is fundamental and often appears in problems involving the substitution method.The natural logarithm emerges in this context due to the properties of the derivative and antiderivative of the \( \ln \) function:- The derivative of \( \ln |u| \) is \( \frac{1}{u} \), making it the perfect candidate for integrals of the form \( \int \frac{1}{u} \, du \).- When integrating functions that result in a log form, it usually indicates a relationship inherent in exponential or rational expressions.In our specific problem, after substituting and transforming the integral, we reached \( \int \frac{du}{u} \), which directly integrates to \( \ln |u| + C \). Upon substituting back \( u = 3 + e^{t} \), the solution becomes \( \ln |3+e^{t}| + C \), showcasing how natural logarithms often simplify complex integration problems.