Using the substitution u = 1+x^2, convert the integral into one involving only the variable u. You do not need to compute the resulting integral.

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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  1. Using the substitution u = 1+x^2, convert the integral into one involving only the variable u. You do not need to compute the resulting integral.

The image displays a mathematical integral commonly encountered in calculus. The integral to be evaluated is:

\[ \int \frac{1}{1 + x^2} \, dx \]

This integral represents the antiderivative of the function \(\frac{1}{1 + x^2}\) with respect to \(x\).

To solve this integral, we recognize that it fits the form of the standard integral formula:

\[ \int \frac{1}{1 + x^2} \, dx = \arctan(x) + C \]

where \(\arctan(x)\) is the inverse tangent function, and \(C\) is the constant of integration. 

This result follows from the fact that the derivative of \(\arctan(x)\) is \(\frac{ d }{ dx } \left( \arctan(x) \right) = \frac{1}{1 + x^2}\).
Transcribed Image Text:The image displays a mathematical integral commonly encountered in calculus. The integral to be evaluated is: \[ \int \frac{1}{1 + x^2} \, dx \] This integral represents the antiderivative of the function \(\frac{1}{1 + x^2}\) with respect to \(x\). To solve this integral, we recognize that it fits the form of the standard integral formula: \[ \int \frac{1}{1 + x^2} \, dx = \arctan(x) + C \] where \(\arctan(x)\) is the inverse tangent function, and \(C\) is the constant of integration. This result follows from the fact that the derivative of \(\arctan(x)\) is \(\frac{ d }{ dx } \left( \arctan(x) \right) = \frac{1}{1 + x^2}\).
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