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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.3.45
Chapter 7, Problem 7.3.45

Evaluate the integrals in Exercises 33–54.
∫ (e^(1/x) / x²) dx

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1
Identify the integral to solve: \(\int \frac{e^{1/x}}{x^{2}} \, dx\).
Look for a substitution that simplifies the exponent and the denominator. Notice that the exponent is \(\frac{1}{x}\), so let \(u = \frac{1}{x}\).
Compute the differential \(du\) in terms of \(dx\): Since \(u = x^{-1}\), then \(du = -x^{-2} \, dx\), or equivalently, \(-du = x^{-2} \, dx\).
Rewrite the integral in terms of \(u\) and \(du\): Substitute \(e^{1/x} = e^{u}\) and \(x^{-2} \, dx = -du\), so the integral becomes \(\int e^{u} (-du) = -\int e^{u} \, du\).
Integrate with respect to \(u\): The integral of \(e^{u}\) is \(e^{u}\), so the integral becomes \(-e^{u} + C\). Finally, substitute back \(u = \frac{1}{x}\) to express the answer in terms of \(x\).

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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, which transforms the integral into a simpler form. This technique is especially useful when the integral contains a composite function.
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Substitution With an Extra Variable

Chain Rule in Reverse

The chain rule in differentiation helps find derivatives of composite functions. In integration, recognizing the chain rule in reverse allows us to identify integrals where the integrand is a function and its derivative, enabling easier substitution and integration.
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Intro to the Chain Rule

Exponential Functions with Variable Exponents

Exponential functions with variable exponents, such as e^(1/x), require careful handling during integration. Understanding how to differentiate and integrate these functions, often through substitution, is essential to solving integrals involving such expressions.
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Related Practice
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