Textbook Question
9–40. Integration by parts Evaluate the following integrals using integration by parts.
20. ∫ sin⁻¹(x) dx
Ch. 8 - Integration Techniques
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 8, Problem 8.2.53
50-53. Reduction Formulas Use integration by parts to derive the following reduction formulas:
53. ∫ lnⁿ(x) dx = x lnⁿ(x) - n ∫ lnⁿ⁻¹(x) dx
Verified step by step guidance1
Identify the integral to solve: \(\int \ln^{n}(x) \, dx\), where \(n\) is a positive integer power of the natural logarithm function.
Choose parts for integration by parts. Let \(u = \ln^{n}(x)\) and \(dv = dx\). Then, compute \(du\) and \(v\):
- Since \(u = \ln^{n}(x)\), use the chain rule to find \(du = n \ln^{n-1}(x) \cdot \frac{1}{x} \, dx\).
- Since \(dv = dx\), integrate to get \(v = x\).
Apply the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). Substitute the expressions for \(u\), \(v\), and \(du\):
\(\int \ln^{n}(x) \, dx = x \ln^{n}(x) - \int x \cdot n \ln^{n-1}(x) \cdot \frac{1}{x} \, dx\).
Simplify the integral inside the subtraction:
\(\int x \cdot n \ln^{n-1}(x) \cdot \frac{1}{x} \, dx = n \int \ln^{n-1}(x) \, dx\).
Rewrite the expression to obtain the reduction formula:
\(\int \ln^{n}(x) \, dx = x \ln^{n}(x) - n \int \ln^{n-1}(x) \, dx\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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 essential for deriving reduction formulas involving powers of logarithmic functions.
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Integration by Parts for Definite Integrals
Reduction Formulas
Reduction formulas express an integral involving a power or parameter in terms of a similar integral with a lower power or simpler parameter. They simplify complex integrals step-by-step, making it easier to evaluate integrals like ∫lnⁿ(x) dx by relating them to ∫lnⁿ⁻¹(x) dx.
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5:59Recursive Formulas
Properties of the Natural Logarithm Function
Understanding the natural logarithm function ln(x) and its derivatives is crucial. Since d/dx[ln(x)] = 1/x, this property helps in choosing appropriate parts for integration by parts and manipulating integrals involving powers of ln(x).
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Related Practice
Textbook Question
7–64. Integration review Evaluate the following integrals.
20. ∫ eˣ (1 + eˣ)⁹ (1 - eˣ) dx
Textbook Question
5. What is a reduction formula?
Textbook Question
7–40. Table look-up integrals Use a table of integrals to evaluate the following indefinite integrals. Some of the integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table.
24. ∫ dt / √(1 + 4eᵗ)
Textbook Question
What are the two general ways in which an improper integral may occur?
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Textbook Question
7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.
54. ∫ y⁴/(1 + y²) dy