Textbook Question
7–64. Integration review Evaluate the following integrals.
53. ∫ eˣ sec(eˣ + 1) dx
Ch. 8 - Integration Techniques
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 8, Problem 8.5.23
23-64. Integration Evaluate the following integrals.
23. ∫ [3 / ((x - 1)(x + 2))] dx
Verified step by step guidance1
Start by expressing the integrand \( \frac{3}{(x - 1)(x + 2)} \) as a sum of partial fractions. Assume it can be written as \( \frac{A}{x - 1} + \frac{B}{x + 2} \), where \(A\) and \(B\) are constants to be determined.
Multiply both sides of the equation by the denominator \( (x - 1)(x + 2) \) to clear the fractions, resulting in \( 3 = A(x + 2) + B(x - 1) \).
Expand the right-hand side to get \( 3 = A x + 2A + B x - B \), then group like terms: \( 3 = (A + B) x + (2A - B) \).
Set up a system of equations by equating the coefficients of corresponding powers of \(x\) on both sides. Since the left side has no \(x\) term, the coefficient of \(x\) must be zero, and the constant term must be 3. So, \( A + B = 0 \) and \( 2A - B = 3 \).
Solve the system for \(A\) and \(B\), then rewrite the integral as \( \int \left( \frac{A}{x - 1} + \frac{B}{x + 2} \right) dx \). Finally, integrate each term separately using the formula \( \int \frac{1}{x - c} dx = \ln|x - c| + C \).
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4mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to break down a complex rational function into simpler fractions that are easier to integrate. It involves expressing the integrand as a sum of fractions with linear or quadratic denominators, allowing straightforward integration of each term.
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Integration of Rational Functions
Integrating rational functions often requires rewriting the integrand into simpler parts, such as partial fractions. Once decomposed, each term can be integrated using basic integral formulas, typically involving logarithmic functions for linear denominators.
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Logarithmic Integration
When integrating functions of the form 1/(ax + b), the result is a logarithmic function ln|ax + b|/a plus a constant. This concept is essential for integrating the terms obtained after partial fraction decomposition with linear denominators.
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Related Practice
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