Problem 114 - Calculus and its applications

Question

Evaluate. $$ \int \frac{e^{-m x}}{1-a e^{-m x}} d x $$

Step-by-Step Solution

Answer

\( \frac{1}{a m} \ln|1 - a e^{-mx}| + C \).

Step 1: Substitution

Begin by using the substitution method. Let \( u = 1 - a e^{-mx} \). Then, find \( du \). The derivative of \( u \) with respect to \( x \) is \( du = a m e^{-mx} \, dx \). Rearrange this to express \( dx \) in terms of \( du \): \( dx = \frac{du}{a m e^{-mx}} \).

Step 2: Simplifying the Integrand

Substitute \( u \) and \( dx \) into the integral. The integral becomes \( \int \frac{e^{-mx}}{u} \cdot \frac{du}{a m e^{-mx}} \). Simplify this to \( \int \frac{1}{a m u} \, du \), which further simplifies to \( \frac{1}{a m} \int \frac{1}{u} \, du \).

Step 3: Integrate Using Logarithmic Rule

Now, integrate \( \int \frac{1}{u} \, du \). This is a standard integral that results in \( \ln|u| + C \), where \( C \) is the constant of integration. So, \( \frac{1}{a m} \int \frac{1}{u} \, du = \frac{1}{a m} \ln|u| + C \).

Step 4: Substitution Back to Original Variable

Replace \( u \) with its original expression \( 1 - a e^{-mx} \). The solution of the integral is \( \frac{1}{a m} \ln|1 - a e^{-mx}| + C \).

Key Concepts

Substitution Method

The substitution method is a powerful technique in integration. It helps to transform a more complicated integral into a simpler one. The key idea is to identify a part of the integrand that can be substituted with a single variable. This simplification often involves finding a new variable, typically called \( u \).In our exercise, we look at the integral \( \int \frac{e^{-m x}}{1-a e^{-m x}} d x \), where the substitution \( u = 1 - a e^{-mx} \) has been used. The rationale behind choosing this substitution is to simplify the denominator, allowing a clean path forward during integration.

• Find \( u \) such that it significantly simplifies the problem.
• Determine the derivative \( du \) of \( u \) with respect to \( x \).
• Rewrite \( dx \) in terms of \( du \), providing a full replacement.

By substituting \( u \) and replacing \( dx \) in terms of \( du \), the integration problem changes form, often revealing a much simpler integral.

Logarithmic Integration

Logarithmic integration is an essential technique for functions that have the form of reciprocal expressions. The integral \( \int \frac{1}{u} \, du \) is a classic example, resulting in the natural logarithm function \( \ln|u| \). This form appears frequently and is a fundamental result of calculus.Returning to the exercise, once we substituted \( u = 1 - a e^{-mx} \), and simplified the integrand to \( \int \frac{1}{u} \, du \), the integral could be directly approached using logarithmic integration.

• Recognize when the integrand fits the form that applies to logarithmic rules.
• The integral of \( \frac{1}{u} \, du \) directly gives \( \ln|u| + C \), where \( C \) is the constant of integration.
• Logarithmic results are central for simplifying problems involving inverses and growth measurements.

Integration with Exponential Functions

Integration involving exponential functions requires careful handling to maintain the constant increases or decreases characteristic of these functions. Exponential functions, like \( e^{-mx} \), often appear in various scientific and engineering contexts, necessitating methods for their integration.In the exercise, an exponential function is a part of the initially more complex quotient. Using substitution helped manage this efficiently, but understanding the behavior of exponential terms provides additional insight.

• Remember that integration may sometimes require multiplicative constants, as modifying factors accompany derivatives.
• Substitution often helps when the exponential is part of more significant expressions.
• Recognize patterns and factor changes due to the derivative of \( e^{x} \) being itself times a constant.

This approach allows the intrinsic properties of exponentials to be utilized effectively, achieving a more straightforward resolution of the integral.