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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.2.28
Chapter 8, Problem 8.2.28

Evaluate the integrals in Exercises 25–30 by using a substitution prior to integration by parts.
∫ ln(x + x²) dx

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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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

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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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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Integration 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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Logarithms Introduction
Related Practice
Textbook Question

In Exercises 35–68, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer.

∫ from 1 to ∞ of ((1 / (e^x - 2^x)) dx)

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Textbook Question

Use any method to evaluate the integrals in Exercises 15–38. Most will require trigonometric substitutions, but some can be evaluated by other methods.

∫ (x dx) / (25 + 4x²)

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Textbook Question

Evaluate the integrals in Exercises 31–56. Some integrals do not require integration by parts.

∫ x⁵ e³ˣ dx

Textbook Question

Use any method to evaluate the integrals in Exercises 65–70.

∫ sin³(x) / cos⁴(x) dx

Textbook Question

In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.

∫ x^2 √(2x - x^2) dx

Textbook Question

Solve the initial value problems in Exercises 53–56 for y as a function of x.

√(x² - 9) (dy/dx) = 1, where x > 3, y(5) = ln 3