Textbook Question
Evaluate the integrals in Exercises 39–56.
56. ∫sec(x)dx/√(ln(sec(x)+tan(x)))
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- 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
Chapter 7, Problem 7.3.95
Evaluate the integrals in Exercises 87–96.
95. ∫₂⁴ x^(2x) (1 + ln x) dx
Verified step by step guidance1
Recognize that the integrand is of the form \(x^{2x} (1 + \ln x)\), which suggests a function multiplied by its derivative or a derivative of a product involving \(x^{2x}\).
Rewrite the integrand by expressing \(x^{2x}\) in terms of the exponential function: \(x^{2x} = e^{2x \ln x}\).
Differentiate \(x^{2x}\) with respect to \(x\) using the chain rule: \(\frac{d}{dx} x^{2x} = \frac{d}{dx} e^{2x \ln x} = e^{2x \ln x} \cdot \frac{d}{dx} (2x \ln x)\).
Calculate \(\frac{d}{dx} (2x \ln x)\) using the product rule: \(\frac{d}{dx} (2x \ln x) = 2 \ln x + 2\).
Notice that the integrand \(x^{2x} (1 + \ln x)\) matches \(\frac{1}{2} \frac{d}{dx} x^{2x}\), so rewrite the integral accordingly and integrate by reversing the differentiation.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Integration by Substitution
Integration by substitution is a technique used to simplify integrals by changing variables. It involves identifying a part of the integrand as a new variable, which transforms the integral into a simpler form. This method is especially useful when the integral contains a function and its derivative.
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Substitution With an Extra Variable
Differentiation of Exponential Functions with Variable Exponents
Functions like x^(2x) involve variable exponents, which require logarithmic differentiation to handle. Understanding how to differentiate and integrate such functions is crucial, as they combine polynomial and exponential behaviors. Recognizing the derivative of the exponent helps in integration.
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6:13Exponential Functions
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links differentiation and integration, allowing evaluation of definite integrals using antiderivatives. After finding an antiderivative of the integrand, you compute its values at the upper and lower limits and subtract to find the integral's value.
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06:11Fundamental Theorem of Calculus Part 1
Related Practice
Textbook Question
In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.
37. ∫(from x²/2 to x²)ln(√t)dt
Textbook Question
Evaluate the integrals in Exercises 87–96.
87. ∫ 5ˣ dx
Textbook Question
In Exercises 13–24, find the derivative of y with respect to the appropriate variable.
15. y = 2√t tanh(√t)
Textbook Question
Evaluate the integrals in Exercises 41–60.
49. ∫(sech(√t)tanh(√t)dt)/√t
Textbook Question
In Exercises 115–126, use logarithmic differentiation or the method in Example 6 to find the derivative of y with respect to the given independent variable.
120. y = x^(sin x)