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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 8, Problema 8.5.40

Evaluate the integrals in Exercises 39–54.
∫ (e⁴t + 2e²t - e^t) / (e²t + 1) dt

Guida verificata passo dopo passo
1
Start by examining the integral \( \int \frac{e^{4t} + 2e^{2t} - e^{t}}{e^{2t} + 1} \, dt \). Notice that the numerator and denominator involve exponential functions with powers related to \( t \).
Make a substitution to simplify the expression. Let \( u = e^{2t} \). Then, compute \( du \) in terms of \( dt \): \( du = 2e^{2t} dt = 2u dt \), so \( dt = \frac{du}{2u} \).
Rewrite the integral in terms of \( u \). Express each term in the numerator and denominator using \( u \): \( e^{4t} = (e^{2t})^2 = u^2 \), \( 2e^{2t} = 2u \), and \( e^{t} = e^{t} \) (which can be expressed as \( e^{t} = (e^{2t})^{1/2} = u^{1/2} \)). Substitute these into the integral and replace \( dt \) with \( \frac{du}{2u} \).
Simplify the resulting integral in terms of \( u \), combining like terms and simplifying the fraction if possible. This should reduce the integral to a rational function of \( u \) and possibly \( u^{1/2} \).
Once simplified, split the integral into simpler parts if needed and integrate each term with respect to \( u \). After integrating, substitute back \( u = e^{2t} \) to express the answer in terms of \( t \).

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