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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
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Tutti i libri di testoHass 15th EditionCh. 7 - Transcendental FunctionsProblema 7.7.69b
Capitolo 7, Problema 7.7.69b

Evaluate the integrals in Exercises 67–74 in terms of
b. natural logarithms.
69. ∫(from 5/4 to 2)dx/(1-x²)

Guida verificata passo dopo passo
1
Recognize that the integral is of the form \( \int \frac{dx}{1 - x^2} \), which can be rewritten using partial fractions since \(1 - x^2 = (1 - x)(1 + x)\).
Set up the partial fraction decomposition: \( \frac{1}{1 - x^2} = \frac{A}{1 - x} + \frac{B}{1 + x} \).
Solve for constants \(A\) and \(B\) by multiplying both sides by \(1 - x^2\) and equating coefficients or substituting convenient values for \(x\).
Rewrite the integral as the sum of two simpler integrals: \( \int \frac{A}{1 - x} dx + \int \frac{B}{1 + x} dx \), and integrate each term separately, recalling that \( \int \frac{1}{a - x} dx = -\ln|a - x| + C \) and \( \int \frac{1}{a + x} dx = \ln|a + x| + C \).
Apply the definite integral limits from \( \frac{5}{4} \) to \( 2 \) to the resulting expression involving natural logarithms, and write the answer in terms of natural logarithms without simplifying the numerical value.

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Definite Integrals

A definite integral calculates the net area under a curve between two specific limits. It is evaluated by finding the antiderivative of the integrand and then applying the Fundamental Theorem of Calculus to subtract the values at the upper and lower bounds.
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Integration of Rational Functions

Integrating rational functions often involves algebraic manipulation such as partial fraction decomposition. This technique breaks a complex fraction into simpler terms that are easier to integrate, especially when the denominator factors into linear or quadratic terms.
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Natural Logarithm in Integration

The natural logarithm function, ln(x), frequently appears as the antiderivative of functions of the form 1/x or similar rational expressions. Recognizing when an integral results in a logarithmic function is key to expressing the solution in terms of natural logarithms.
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Derivative of the Natural Logarithmic Function
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