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(Зх — 1) (x² – 1) A B (x + 1) (x – 1) A = %3D В (Зх — 1) dx (x² – 1) - +C -

(Зх — 1) (x² – 1) A B (x + 1) (x – 1) A = %3D В (Зх — 1) dx (x² – 1) - +C -

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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The image contains a mathematical expression involving partial fraction decomposition and integration. The expression given is: \[ \frac{3x - 1}{x^2 - 1} = \frac{A}{x + 1} + \frac{B}{x - 1} \] This is followed by fields to solve for constants \( A \) and \( B \): - \( A = \) (input field) - \( B = \) (input field) Finally, there is an integral expression to evaluate: \[ \int \frac{3x - 1}{x^2 - 1} \, dx = \text{(input field)} + C \] The purpose of this expression is to find the values of \( A \) and \( B \) that satisfy the equation, and then use them to evaluate the integral. The constant \( C \) represents the constant of integration.
Transcribed Image Text:The image contains a mathematical expression involving partial fraction decomposition and integration. The expression given is: \[ \frac{3x - 1}{x^2 - 1} = \frac{A}{x + 1} + \frac{B}{x - 1} \] This is followed by fields to solve for constants \( A \) and \( B \): - \( A = \) (input field) - \( B = \) (input field) Finally, there is an integral expression to evaluate: \[ \int \frac{3x - 1}{x^2 - 1} \, dx = \text{(input field)} + C \] The purpose of this expression is to find the values of \( A \) and \( B \) that satisfy the equation, and then use them to evaluate the integral. The constant \( C \) represents the constant of integration.
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