Textbook Question
Each of Exercises 25–36 gives a formula for a function y=f(x). In each case, find f^(-1)(x) and identify the domain and range of f^(-1). As a check, show that f(f^(-1)(x))=f^(-1)(f(x))=x.
f(x) = 1/x², x > 0
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.2.41
Evaluate the integrals in Exercises 39–56.
41. ∫2y dy/(y²-25)
Verified step by step guidance1
Identify the integral to solve: \(\int \frac{2y}{y^{2} - 25} \, dy\).
Recognize that the denominator is a quadratic expression that can be factored as \(y^{2} - 25 = (y - 5)(y + 5)\), but before factoring, check if substitution is more straightforward.
Use substitution by letting \(u = y^{2} - 25\), then compute \(du = 2y \, dy\), which matches the numerator times \(dy\) exactly.
Rewrite the integral in terms of \(u\): \(\int \frac{2y}{y^{2} - 25} \, dy = \int \frac{1}{u} \, du\).
Integrate \(\int \frac{1}{u} \, du\) to get \(\ln|u| + C\), then substitute back \(u = y^{2} - 25\) to express the answer in terms of \(y\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Integration of Rational Functions
This involves integrating functions expressed as the ratio of two polynomials. Techniques often include algebraic manipulation such as factoring or partial fraction decomposition to simplify the integral into manageable parts.
Recommended video:
6:04
Intro to Rational Functions
Partial Fraction Decomposition
A method used to break down complex rational expressions into simpler fractions that are easier to integrate. It is especially useful when the denominator can be factored into linear or quadratic terms.
Recommended video:
10:07Partial Fraction Decomposition: Distinct Linear Factors
Substitution Method
A technique where a substitution is made to simplify the integral, often by letting a part of the integrand equal a new variable. This can transform the integral into a basic form that is straightforward to evaluate.
Recommended video:
07:33Euler's Method
Related Practice
Textbook Question
Use l’Hôpital’s rule to find the limits in Exercises 7–52.
39. lim (x → ∞) (ln 2x - ln(x + 1))
Textbook Question
In Exercises 7–26, find the derivative of y with respect to x, t, or θ, as appropriate.
y = ln(e^(θ)/(1+e^θ))
Textbook Question
Evaluate the integrals in Exercises 53–76.
69. ∫dx/((2x-1)√((2x-1)²-4))
1
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Textbook Question
Evaluate the integrals in Exercises 39–56.
52. ∫(from π/4 to π/2)cot(t)dt
1
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Textbook Question
Each of Exercises 25–36 gives a formula for a function y=f(x). In each case, find f^(-1)(x) and identify the domain and range of f^(-1). As a check, show that f(f^(-1)(x))=f^(-1)(f(x))=x.
f(x) = (x + 3) / (x − 2)