Chapter 6: Problem 29
Find the indefinite integral. $$ \int \frac{e^{-x}}{e^{-x}+1} d x $$
Short Answer
Expert verified
\(-\ln|e^{-x} + 1| + C\) is the indefinite integral.
Step by step solution
01
Recognize the Structure
The integral \( \int \frac{e^{-x}}{e^{-x} + 1} \, dx \) involves a fraction where the numerator is \( e^{-x} \) and the denominator is \( e^{-x} + 1 \). This suggests that a substitution may simplify the problem.
02
Choose a Substitution
Let \( u = e^{-x} + 1 \). Then, compute the derivative: \( \frac{du}{dx} = -e^{-x} \) or \( du = -e^{-x} \, dx \). This means \( -du = e^{-x} \, dx \).
03
Substitute into the Integral
Replace \( e^{-x} \, dx \) in the original integral with \( -du \). The integral becomes \( \int \frac{-1}{u} \, du \).
04
Evaluate the New Integral
The integral \( \int \frac{-1}{u} \, du \) simplifies to \( -\ln|u| + C \) using the logarithmic rule for integrals.
05
Substitute Back to Original Variable
Since we previously let \( u = e^{-x} + 1 \), substitute \( u \) back to get \( -\ln|e^{-x} + 1| + C \). This is the integral in terms of \( x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique in calculus used to simplify integrals by transforming them into a form that is easier to solve. The key is to choose a substitution that converts the integral into one involving a simpler function. For the integral \( \int \frac{e^{-x}}{e^{-x} + 1} \, dx \), the structure suggests that substitution could help. By letting \( u = e^{-x} + 1 \), the integral is transformed. The derivative \( \frac{du}{dx} = -e^{-x} \) helps to express \( dx \) in terms of \( du \). Thus, the new integral becomes \( \int \frac{-1}{u} \, du \). Using substitution effectively can simplify complex integrals considerably, allowing you to work with more manageable expressions. Remember, the goal is to simplify the integral into a basic form you know how to solve.
Logarithmic Integration
Logarithmic integration deals specifically with integrals of the form \( \int \frac{1}{x} \, dx \) which result in logarithmic expressions. This is a very useful integration technique especially when dealing with rational functions. In our exercise, after performing the substitution, the integral \( \int \frac{-1}{u} \, du \) arises. This matches the form \( \int \frac{1}{x} \, dx \), except it has a multiplier of \(-1\). Utilizing logarithmic integration, this simplifies to \(-\ln|u| + C\). Knowing the different forms of integrals that lead to a logarithmic result is crucial. It streamlines solving integrals that might not be immediately recognizable, transforming a potentially complex process into straightforward steps.
Integration Techniques
Integration techniques encompass a range of methods developed to solve integrals that may not seem solvable using simple antiderivatives. Understanding which technique to apply can be crucial for efficient problem solving. Common techniques include
- Substitution: Used when the integral can be simplified by a change of variable, which was used in our example.
- Integration by Parts: Useful for products of functions, based on the product rule for derivatives.
- Partial Fractions: Efficient for rational expressions, where the integrand is split into simpler fractions.
- Trigonometric Substitution: Helpful for integrals involving \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\).