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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.6.45

7–84. Evaluate the following integrals.
45. ∫ from 0 to ln 2 [1 / (1 + eˣ)²] dx

Guida verificata passo dopo passo
1
Step 1: Recognize that the integral involves the function \( \frac{1}{(1 + e^x)^2} \). This suggests that substitution might simplify the problem. Look for a substitution that simplifies the denominator \( 1 + e^x \).
Step 2: Let \( u = 1 + e^x \). Then, differentiate \( u \) with respect to \( x \): \( \frac{du}{dx} = e^x \), or equivalently \( du = e^x dx \). Rewrite \( e^x dx \) in terms of \( u \).
Step 3: Substitute \( u \) into the integral. When \( x = 0 \), \( u = 1 + e^0 = 2 \). When \( x = \ln 2 \), \( u = 1 + e^{\ln 2} = 1 + 2 = 3 \). The integral becomes \( \int_{2}^{3} \frac{1}{u^2} \cdot \frac{du}{e^x} \). Since \( e^x = u - 1 \), replace \( e^x \) with \( u - 1 \).
Step 4: Simplify the integral to \( \int_{2}^{3} \frac{1}{u^2(u - 1)} du \). This integral can be solved using partial fraction decomposition. Express \( \frac{1}{u^2(u - 1)} \) as \( \frac{A}{u} + \frac{B}{u^2} + \frac{C}{u - 1} \), and solve for \( A \), \( B \), and \( C \).
Step 5: After finding the coefficients \( A \), \( B \), and \( C \), rewrite the integral as a sum of simpler integrals: \( \int \frac{A}{u} du + \int \frac{B}{u^2} du + \int \frac{C}{u - 1} du \). Evaluate each term separately and apply the limits of integration \( u = 2 \) to \( u = 3 \).

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Definite Integrals

A definite integral calculates the accumulation of a quantity, represented as the area under a curve, between two specified limits. In this case, the integral from 0 to ln 2 indicates that we are interested in the area under the curve of the function 1 / (1 + eˣ)² from x = 0 to x = ln 2.
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Integration Techniques

To evaluate integrals, various techniques can be employed, such as substitution, integration by parts, or recognizing standard forms. For the given integral, a substitution involving the exponential function may simplify the process, allowing for easier integration of the function.
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Exponential functions, such as eˣ, are fundamental in calculus due to their unique properties, including their derivatives and integrals being proportional to the function itself. Understanding how to manipulate and integrate these functions is crucial for solving integrals involving eˣ, as seen in the integrand of the given problem.
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