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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.3.53

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

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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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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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Substitution With an Extra Variable

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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Exponential Functions

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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Intro to Rational Functions