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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 8, Problem 8.2.60a

60. Two Methods
a. Evaluate ∫(x · ln(x²)) dx using the substitution u = x² and evaluating ∫(ln(u)) du.

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1
Start with the integral \( \int x \cdot \ln(x^2) \, dx \). The goal is to use the substitution \( u = x^2 \).
Compute the differential \( du \) from \( u = x^2 \). Since \( \frac{du}{dx} = 2x \), we have \( du = 2x \, dx \), which implies \( x \, dx = \frac{du}{2} \).
Rewrite the integral in terms of \( u \): \( \int x \cdot \ln(x^2) \, dx = \int \ln(u) \cdot x \, dx = \int \ln(u) \cdot \frac{du}{2} = \frac{1}{2} \int \ln(u) \, du \).
Focus on evaluating \( \int \ln(u) \, du \). Use integration by parts where you let \( v = \ln(u) \) and \( dw = du \). Then, \( dv = \frac{1}{u} du \) and \( w = u \).
Apply integration by parts formula: \( \int v \, dw = vw - \int w \, dv \). Substitute the expressions to get \( u \ln(u) - \int u \cdot \frac{1}{u} du = u \ln(u) - \int 1 \, du = u \ln(u) - u + C \).

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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 substituting a part of the integral with a new variable, making the integral easier to evaluate. In this problem, substituting u = x² transforms the integral into one involving ln(u), which is simpler to handle.
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Substitution With an Extra Variable

Properties of Logarithms

Understanding logarithmic properties is essential when dealing with integrals involving ln(x²). The property ln(x²) = 2 ln(x) can simplify expressions before integration. Recognizing how to manipulate logarithmic functions helps in rewriting the integral in a more manageable form.
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Change of Base Property

Integration of Logarithmic Functions

Integrating functions involving logarithms, such as ∫ln(u) du, requires specific techniques like integration by parts. Knowing the formula ∫ln(u) du = u ln(u) - u + C is useful. This concept is crucial for completing the integral after substitution.
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Graphs of Logarithmic Functions