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Arc length: Find the length of the curve y = ln(sec x), 0 ≤ x ≤ π/4.
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Ch. 8 - Techniques of Integration
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 8, Problema 8.2.28
Evaluate the integrals in Exercises 25–30 by using a substitution prior to integration by parts.
∫ ln(x + x²) dx
Guida verificata passo dopo passo1
Start by examining the integral \(\int \ln(x + x^{2}) \, dx\). Notice that the argument of the logarithm can be factored to simplify the expression inside the logarithm.
Factor the expression inside the logarithm: \(x + x^{2} = x(1 + x)\). Use the logarithm property \(\ln(ab) = \ln a + \ln b\) to rewrite the integral as \(\int \ln x + \ln(1 + x) \, dx\).
Split the integral into two separate integrals: \(\int \ln x \, dx + \int \ln(1 + x) \, dx\). This makes the problem easier to handle by dealing with each term individually.
For each integral, use integration by parts. Recall the formula for integration by parts: \(\int u \, dv = uv - \int v \, du\). Choose \(u\) to be the logarithmic function (e.g., \(u = \ln x\) or \(u = \ln(1 + x)\)) and \(dv\) to be \(dx\).
Compute \(du\) and \(v\) for each integral: For \(u = \ln x\), \(du = \frac{1}{x} dx\); for $dv = dx$, \(v = x\). Similarly, for \(u = \ln(1 + x)\), \(du = \frac{1}{1 + x} dx\); for $dv = dx$, \(v = x\). Then apply the integration by parts formula to each integral separately.
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Substitution Method
The substitution method simplifies integrals by changing variables to transform the integral into a more manageable form. It involves choosing a substitution u = g(x) that simplifies the integrand, then rewriting the integral in terms of u and du. This technique is often used to prepare an integral for further methods like integration by parts.
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07:33
Euler's Method
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. Choosing u and dv wisely is crucial to simplify the integral effectively.
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Percorso guidato
06:18Integration by Parts for Definite Integrals
Logarithmic Integration
Integrals involving logarithmic functions often require special techniques, such as substitution or integration by parts, because logarithms do not have elementary antiderivatives in simple form. Recognizing how to manipulate the argument of the logarithm can simplify the integral and make it solvable.
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7:30Logarithms Introduction
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