Thomas' Calculus 13th Edition

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

Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 4



Work Step by Step

Put u= $x^{4}+1$. $\frac{du}{dx}= 4x^{3}$ or dx=$\frac{du}{4x^{3}}$ Then, I= $\int(\frac{4x^{3}}{u^{2}}\times\frac{du}{4x^{3}})=\int u^{-2}du$ =$ \frac{u^{-2+1}}{-2+1}+C= \frac{-1}{u}+C$. Substituting the value of u, we have $\int \frac{4x^{3}}{(x^{4}+1)^{2}}dx=-\frac{1}{x^{4}+1}$+C
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