Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 347: 83



Work Step by Step

To evaluate the integral $$\int_0^1\frac{e^z+1}{e^z+z}dz$$ we will use substitution $e^z+z=t$ which gives us $(e^z+1)dz=dt$. The integration bounds would be: for $z=0$ we have $t=1$ and for $z=1$ we have $t=e+1$. Putting this into the integral we get: $$\int_0^1\frac{e^z+1}{e^z+z}dz=\int_1^{e+1}\frac{1}{t}dx=\left.\ln|t|\right|_1^{e+1}=\ln(e+1)-\ln1=\ln(e+1)$$
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