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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.3.53

Evaluate the integrals in Exercises 33–54.
53. ∫ (e^r / (1 + e^r)) dr

Verified step by step guidance
1
Recognize that the integral is of the form \(\int \frac{e^r}{1 + e^r} \, dr\). This suggests a substitution involving the denominator \(1 + e^r\) because its derivative appears in the numerator.
Let \(u = 1 + e^r\). Then, compute the derivative \(\frac{du}{dr} = e^r\), which means \(du = e^r \, dr\).
Rewrite the integral in terms of \(u\): since \(e^r \, dr = du\), the integral becomes \(\int \frac{1}{u} \, du\).
Integrate \(\int \frac{1}{u} \, du\), which is a standard integral resulting in \(\ln|u| + C\).
Substitute back \(u = 1 + e^r\) to express the answer in terms of the original variable \(r\), giving \(\ln|1 + e^r| + C\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration by Substitution

Integration by substitution is a method used to simplify integrals by changing variables. It involves identifying a part of the integrand as a new variable, which transforms the integral into a simpler form. This technique is especially useful when the integral contains a composite function.
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Exponential Functions

Exponential functions have the form e^x, where e is Euler's number. They exhibit unique properties such as their derivative and integral being proportional to themselves. Understanding how to manipulate and integrate exponential functions is essential for solving integrals involving e^r.
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Rational Functions and Simplification

Rational functions are ratios of polynomials or expressions involving variables. Simplifying such functions before integration can make the process easier. Recognizing patterns in the numerator and denominator, such as derivatives of the denominator appearing in the numerator, helps in choosing the right substitution.
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