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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 8.2.54

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.
∫ (xe^x) / (x + 1)² dx

Guida verificata passo dopo passo
1
Recognize that the integral is of the form \(\int \frac{x e^{x}}{(x+1)^2} \, dx\). Since the denominator is \((x+1)^2\), consider using a substitution to simplify the expression.
Let \(u = x + 1\). Then, \(x = u - 1\) and $dx = du$. Rewrite the integral in terms of \(u\): \(\int \frac{(u - 1) e^{u - 1}}{u^2} \, du\).
Rewrite the exponential term as \(e^{u - 1} = e^{u} e^{-1} = \frac{e^{u}}{e}\). So the integral becomes \(\frac{1}{e} \int \frac{(u - 1) e^{u}}{u^2} \, du\).
Split the integral into two parts: \(\frac{1}{e} \int \frac{u e^{u}}{u^2} \, du - \frac{1}{e} \int \frac{e^{u}}{u^2} \, du\), which simplifies to \(\frac{1}{e} \int \frac{e^{u}}{u} \, du - \frac{1}{e} \int \frac{e^{u}}{u^2} \, du\).
At this point, consider using integration by parts on the integral \(\int \frac{e^{u}}{u^2} \, du\) to simplify it further, choosing appropriate functions for \(u\) and \(dv\) to reduce the power of \(u\) in the denominator.

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Integration by Parts

Integration by parts is a technique based on the product rule for differentiation. It transforms the integral of a product of functions into simpler integrals, using the formula ∫u dv = uv - ∫v du. This method is useful when the integrand is a product of algebraic and exponential or logarithmic functions.
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The substitution method simplifies integrals by changing variables to reduce the integral into a more manageable form. It involves identifying a part of the integrand as a new variable, which helps in rewriting the integral in terms of this variable, often making the integral easier to evaluate.
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Rational Functions and Simplification

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