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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.6.73

Evaluate the integrals in Exercises 53–76.
73. ∫(from 0 to ln√3) e^x dx/(1+e^(2x))

Guida verificata passo dopo passo
1
Recognize the integral to evaluate: \(\int_0^{\ln \sqrt{3}} \frac{e^x}{1 + e^{2x}} \, dx\).
Make a substitution to simplify the integral. Let \(u = e^x\). Then, the differential is \(du = e^x \, dx\), which means \(e^x \, dx = du\).
Rewrite the integral in terms of \(u\). The limits change as follows: when \(x = 0\), \(u = e^0 = 1\); when \(x = \ln \sqrt{3}\), \(u = e^{\ln \sqrt{3}} = \sqrt{3}\). The integral becomes \(\int_1^{\sqrt{3}} \frac{1}{1 + u^2} \, du\).
Recognize that \(\int \frac{1}{1 + u^2} \, du\) is the standard integral for \(\arctan u + C\). So, the integral evaluates to \(\arctan u\) evaluated from \(1\) to \(\sqrt{3}\).
Apply the Fundamental Theorem of Calculus by substituting the limits back into \(\arctan u\) to express the definite integral as \(\arctan(\sqrt{3}) - \arctan(1)\).

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