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Calculus
cln 2 0 ex √e²x - 1 dx

cln 2 0 ex √e²x - 1 dx

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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How to find the integral?
The image shows the following integral expression: \[ \int_{0}^{\ln 2} \frac{e^x}{\sqrt{e^{2x} - 1}} \, dx \] This integral is to be evaluated from the lower limit \(x = 0\) to the upper limit \(x = \ln 2\). The integrand is the function \(\frac{e^x}{\sqrt{e^{2x} - 1}}\). Here, \(e^x\) is the exponential function, and the expression under the square root is \(e^{2x} - 1\).
Transcribed Image Text:The image shows the following integral expression: \[ \int_{0}^{\ln 2} \frac{e^x}{\sqrt{e^{2x} - 1}} \, dx \] This integral is to be evaluated from the lower limit \(x = 0\) to the upper limit \(x = \ln 2\). The integrand is the function \(\frac{e^x}{\sqrt{e^{2x} - 1}}\). Here, \(e^x\) is the exponential function, and the expression under the square root is \(e^{2x} - 1\).
Expert Solution
Step 1: To find

integral subscript 0 superscript ln open parentheses 2 close parentheses end superscript fraction numerator e to the power of x over denominator square root of e to the power of 2 x end exponent minus 1 end root end fraction d x

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