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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 8.1.34

The integrals in Exercises 1–44 are in no particular order. Evaluate each integral using any algebraic method, trigonometric identity, or substitution you think is appropriate.
∫ e^(z + eᶻ) dz

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1
Identify the integral to solve: \(\int e^{z + e^{z}} \, dz\).
Rewrite the integrand by separating the exponent: \(e^{z + e^{z}} = e^{z} \cdot e^{e^{z}}\).
Consider a substitution to simplify the integral. Let \(u = e^{z}\), so that \(\frac{du}{dz} = e^{z} = u\), which implies \(dz = \frac{du}{u}\).
Rewrite the integral in terms of \(u\): \(\int e^{z} \cdot e^{e^{z}} \, dz = \int u \cdot e^{u} \cdot \frac{du}{u} = \int e^{u} \, du\).
Integrate \(\int e^{u} \, du\) to get \(e^{u} + C\), then substitute back \(u = e^{z}\) to express the answer in terms of \(z\).

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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 have the form e^u, where u is a function of the variable. Understanding how to differentiate and integrate exponential functions, especially when the exponent is itself a function, is crucial. Recognizing the chain rule in reverse helps in integrating such expressions.
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Recognizing Composite Functions in Integrals

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