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Evaluate the definite integral.$\int_{0}^{1} \frac{e^{z}+1}{e^{z}+z} d z$
$\ln (e+1)$
Calculus 1 / AB
Calculus 2 / BC
Chapter 5
Integrals
Section 4
The Substitution Rule
Integration Techniques
Missouri State University
Oregon State University
Boston College
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this question asks us to evaluate the integral we know that are you Is each Z plus z Therefore are d'you is eat Z plus one cause the derivative of Z is simply want Deasy Therefore we have the bounds from E plus one We changed it from one and then we have one on the bottom We changed it from zero of d'you over you which we know gives us natural log of you from one t plus one. Therefore we can plug in Remember, the upper bound is always gonna be subtracting The lower bound natural above one is simply zero There for the answer is natural log of e plus one
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