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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 7.7.69a

Evaluate the integrals in Exercises 67–74 in terms of
a. inverse hyperbolic functions.
69. ∫(from 5/4 to 2)dx/(1-x²)

Verified step by step guidance
1
Recognize that the integral \( \int \frac{dx}{1 - x^2} \) can be rewritten using partial fractions because the denominator factors as \( (1 - x)(1 + x) \).
Express the integrand as partial fractions: \( \frac{1}{1 - x^2} = \frac{A}{1 - x} + \frac{B}{1 + x} \). Solve for constants \( A \) and \( B \).
Recall that the integral of \( \frac{1}{1 - x^2} \) can also be expressed in terms of inverse hyperbolic functions, specifically \( \tanh^{-1}(x) \), since \( \frac{d}{dx} \tanh^{-1}(x) = \frac{1}{1 - x^2} \) for \( |x| < 1 \).
Set up the definite integral from \( x = \frac{5}{4} \) to \( x = 2 \) and write the antiderivative in terms of \( \tanh^{-1}(x) \) or equivalently \( \frac{1}{2} \ln \left| \frac{1 + x}{1 - x} \right| \).
Evaluate the antiderivative at the upper and lower limits and subtract to find the value of the definite integral in terms of inverse hyperbolic functions.

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Key Concepts

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

Definite Integrals

A definite integral calculates the net area under a curve between two specific limits. It is represented as ∫ from a to b of f(x) dx, where a and b are the lower and upper bounds. Evaluating definite integrals involves finding the antiderivative and then applying the Fundamental Theorem of Calculus.
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Inverse Hyperbolic Functions

Inverse hyperbolic functions, such as arsinh, arcosh, and artanh, are the inverses of hyperbolic functions. They often appear as antiderivatives of rational functions involving expressions like 1 - x². Recognizing when to express integrals in terms of these functions simplifies evaluation and interpretation.
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Integration of Rational Functions Involving 1 - x²

Integrals of the form ∫ dx/(1 - x²) can be decomposed using partial fractions or recognized as derivatives of inverse hyperbolic functions. Specifically, ∫ dx/(1 - x²) relates to the inverse hyperbolic tangent function, artanh(x), which helps in expressing the integral in closed form.
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