Problem 42 - Applied Calculus

Question

Find each indefinite integral by the substitution method or state that it cannot be found by our substitution formulas. $$ \int \frac{e^{3 x}}{e^{3 x}-1} d x $$

Step-by-Step Solution

Answer

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

Step 1: Identify the substitution variable

Examine the integral \( \int \frac{e^{3x}}{e^{3x}-1} \, dx \) and look for an appropriate substitution. Notice that the expression in the denominator \( e^{3x} - 1 \) suggests using \( u = e^{3x} - 1 \).

Step 2: Compute the derivative of the substitution

Differentiate the substitution \( u = e^{3x} - 1 \) with respect to \( x \). This gives \( \frac{du}{dx} = 3e^{3x} \) or \( du = 3e^{3x} \, dx \). Hence, \( dx = \frac{du}{3e^{3x}} \).

Step 3: Substitute and simplify the integral

Substitute \( u = e^{3x} - 1 \) and \( dx = \frac{du}{3e^{3x}} \) into the integral:\[ \int \frac{e^{3x}}{e^{3x}-1} \, dx = \int \frac{e^{3x}}{u} \cdot \frac{du}{3e^{3x}} = \frac{1}{3} \int \frac{1}{u} \, du \].

Step 4: Integrate with respect to the substitution variable

The integral \( \frac{1}{3} \int \frac{1}{u} \, du \) can be evaluated as:\[ \frac{1}{3} \ln|u| + C \] where \( C \) is the constant of integration.

Step 5: Substitute back the original variable

Replace \( u \) with \( e^{3x} - 1 \) to get the result in terms of \( x \):\[ \frac{1}{3} \ln|e^{3x} - 1| + C \].

Key Concepts

Substitution Method

In calculus, the substitution method is a key technique used for simplifying the process of integration. It's especially useful when dealing with complex expressions, like products or compositions of functions.

The goal of substitution is to transform the integral into a simpler form that is easier to evaluate. In essence, you "change variables" by substituting a part of the integrand (the function being integrated) with a single variable.

• First, identify a "substitution variable". This is often a component of the integrand, such as a polynomial, trigonometric function, or an exponential function like in our example.
• Next, you compute the derivative of this substitution variable with respect to the original variable, allowing you to replace elements of the integral in terms of the new variable.
• Perform the substitution and simplify the integral accordingly. Ultimately, this often converts the problem into one that involves an elementary function, making it much easier to solve.
• After integrating in terms of the new variable, remember to convert back to the original variable in the final step.

By leveraging substitution, problems that initially seem intractable can become manageable, revealing the integral's true form.

Integration Techniques

Integrating complex functions may require different strategies or techniques. The substitution method is one, but there are many others depending on the type and structure of the function you are dealing with:

• Integration by Parts: Useful for products of functions, particularly involving polynomials and exponentials.
• Partial Fraction Decomposition: Helpful for rational functions with polynomials in the numerator and denominator.
• Trigonometric Integrals: Involve using trigonometric identities to simplify expressions that contain trigonometric functions.

Selecting the appropriate method is key. In the given exercise, substitution helps simplify an otherwise complex integration involving exponential functions.

Each technique has its own application and transforms the integral into a more solvable form. Mastery of these techniques allows you to tackle a wide range of integration problems successfully.

Exponential Functions

Exponential functions are a vital concept in calculus due to their distinctive properties. They are functions of the form \( e^{kx} \), where \( e \) is the base of the natural logarithm, approximately 2.71828, and \( k \) is a constant.

• Exponential functions grow or decay at a rate proportional to their current value, making them crucial in modeling real-world phenomena such as compound interest, population growth, and radioactive decay.
• In the context of integration, exponential functions often appear in compound forms, leading to complex integrands that require advanced techniques like substitution or integration by parts.
• These functions have a unique property where the derivative and the integral of \( e^x \) are both \( e^x \). This property simplifies many calculus problems.

In our example, understanding these properties helped identify \( e^{3x} - 1 \) as a potential substitution, simplifying the integration process. Grasping exponential functions and their integration prepares you to approach a wider array of problems effectively.