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Magazine Chapter 7: Problem 26
Evaluate the integrals. $$\int \frac{d x}{1+e^{x}}$$
Short Answer
Expert verified
The integral evaluates to \( \ln (1 + e^{-x}) + C \).
Step by step solution
01
Identify the Integral Type
The given integral is \( \int \frac{dx}{1+e^{x}} \). This is a rational function where the denominator involves an exponential function. We will use a substitution method to simplify this integral.
02
Choose an Appropriate Substitution
Let's choose the substitution \( u = 1 + e^{x} \). This leads to \( du = e^{x} \, dx \). Since \( e^{x} = u - 1 \), we have \( du = (u - 1) \, dx \). Thus, \( dx = \frac{du}{u - 1} \).
03
Rewrite the Integral
Substitute into the integral: \( \int \frac{dx}{1+e^{x}} = \int \frac{\frac{du}{u-1}}{u} \). This simplifies to \( \int \frac{du}{u(u-1)} \).
04
Use Partial Fraction Decomposition
Since the integrand is a rational function of the form \( \frac{1}{u(u - 1)} \), we decompose it into partial fractions: \( \frac{1}{u(u - 1)} = \frac{A}{u} + \frac{B}{u-1} \). Solving for \( A \) and \( B \), we equate \( 1 = A(u-1) + Bu \) and find \( A = 1 \) and \( B = -1 \).
05
Integrate the Partial Fractions
Substituting back into the integral, we get \( \int \left( \frac{1}{u} - \frac{1}{u - 1} \right) du \). We integrate term by term: \( \int \frac{1}{u} \, du - \int \frac{1}{u-1} \, du = \ln |u| - \ln |u-1| + C \).
06
Back-Substitute for Original Variable
Re-substitute \( u = 1 + e^{x} \) into our integrated expression: \( \ln |1 + e^{x}| - \ln |e^{x}| + C \). Simplify to \( \ln \left| \frac{1 + e^{x}}{e^{x}} \right| + C = \ln \left( 1 + e^{-x} \right) + C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Function
A rational function is any function that can be expressed as the ratio of two polynomials. In simple terms, it's a fraction where both the numerator and the denominator are polynomials.
- For example, the function \( f(x) = \frac{2x + 3}{x^2 - x} \) is a rational function because both the numerator \(2x + 3\) and the denominator \(x^2 - x\) are polynomials.
- Rational functions can have domains that exclude certain values where the denominator becomes zero, creating undefined points or vertical asymptotes.
Substitution Method
The substitution method is a powerful technique used in calculus to simplify integrals. It involves changing the variable of integration to make the integral easier to evaluate.
- This method works well when the integrand, the function to be integrated, can be transformed into a simpler form.
- For example, with the integral \( \int \frac{dx}{1+e^{x}} \), the substitution \( u = 1 + e^{x} \) helps to transform the function into a simpler rational form.
- Choose a substitution \( u = g(x) \) so that the differential \( du = g'(x) \, dx \) simplifies the integral.
- Rewrite the original integral in terms of \( u \) and \( du \), making it easier to solve.
- After integrating with respect to \( u \), substitute back the original variable to express the solution in terms of the original variable of integration.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to break down complex rational expressions into simpler fractions that are easier to integrate or manipulate.
- This technique is particularly useful when dealing with rational functions whose denominators are factorizable.
- For an example like \( \int \frac{1}{u(u-1)} \), it can be decomposed into simpler fractions such as \( \frac{A}{u} + \frac{B}{u-1} \).
- Firstly, assume a decomposition form based on the factors of the denominator.
- Set up an equation that equates the original numerator with the sum of the numerators of these partial fractions, adjusted for the missing factors.
- Solve the resulting linear equation to find the unknown coefficients, such as \( A \) and \( B \).
- Once decomposed, each partial fraction can be integrated individually, allowing for simpler calculation beyond otherwise complex integration problems.
