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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
Chapter 7, Problem 7.PE.45

Evaluate the integrals in Exercises 31–78.
45. ∫(ln x)^(-3)/x dx

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1
Identify the integral to solve: \(\int \frac{(\ln x)^{-3}}{x} \, dx\).
Recognize that the integral involves a composite function with \(\ln x\) inside and a factor of \(\frac{1}{x}\) outside, which suggests using substitution.
Let \(u = \ln x\). Then, compute the differential: \(du = \frac{1}{x} \, dx\), which means \(dx = x \, du\) but since \(du = \frac{1}{x} dx\), we can rewrite \(\frac{1}{x} dx\) as \(du\).
Rewrite the integral in terms of \(u\): \(\int u^{-3} \, du\).
Integrate \(u^{-3}\) using the power rule for integrals: \(\int u^{n} \, du = \frac{u^{n+1}}{n+1} + C\) for \(n \neq -1\). Here, \(n = -3\), so the integral becomes \(\frac{u^{-2}}{-2} + C\).

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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 method used to simplify integrals by changing variables. It involves identifying a part of the integrand as a new variable, differentiating it, and rewriting the integral in terms of this variable to make it easier to solve.
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Properties of Logarithmic Functions

Understanding the natural logarithm function, ln(x), is essential, including its domain (x > 0) and derivative (1/x). Recognizing how powers of ln(x) behave helps in manipulating and integrating expressions involving logarithms.
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Integration of Power Functions

Integrating functions of the form u^n, where n is a real number, follows the power rule for integration. This rule states that ∫u^n du = u^(n+1)/(n+1) + C, provided n ≠ -1, which is crucial when integrating expressions like (ln x)^(-3).
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