Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 40


$-\frac{1}{9 }(x^3+3x)^{-3}+c.$

Work Step by Step

Since $u=x^3+3x$, then $du=(3x^2+3)dx=3(x^2+1)dx$, and hence, we have $$ \int \frac{x^2+1}{(x^3+3x)^4}dx= \frac{1}{3}\int \frac{du}{u^4}=-\frac{1}{9 }u^{-3}+c\\ =-\frac{1}{9 }(x^3+3x)^{-3}+c. $$
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