Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 15: Multiple Integrals - Practice Exercises - Page 933: 11


$$\dfrac{\ln (17)}{4}$$

Work Step by Step

We calculate the integral as follows: $$ \int_{0}^{8} \int_{\sqrt[3] x}^{2} \dfrac{1}{y^4+1} \ dy \ dx =\int_{0}^{2} \int_{0}^{y^3} \dfrac{1}{y^4+1} \ dx \ dy \\= \dfrac{1}{4}\int_{0}^{2} \dfrac{4y^3}{y^4+1} \ dy$$ Set $y^4+1=a \implies 4y^3 dy = da$ Now, $$ \dfrac{1}{4}\int_{0}^{2} \dfrac{4y^3}{y^4+1} dy= \int_{0}^2 \dfrac{a^{-1}}{4} da\\=(\dfrac{1}{4})\times [\ln u]_0^2 \\=\dfrac{\ln (17)}{4}$$
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