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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
8장, 문제 8.2.54

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

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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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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.
추천 영상:
06:18
Integration by Parts for Definite Integrals

Substitution Method

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.
추천 영상:
07:33
Euler's Method

Rational Functions and Simplification

Rational functions are ratios of polynomials, and simplifying them can involve algebraic manipulation such as factoring or dividing polynomials. Recognizing when to simplify the integrand before applying integration techniques can make the integral easier to solve and sometimes avoid more complex methods like integration by parts.
추천 영상:
6:04
Intro to Rational Functions