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: 55


$-\frac{1}{24} (3 x^{4}+9 x^{2})^{-4}+c.$

Work Step by Step

We have \begin{align*} \int \frac{\left(2 x^{3}+3 x\right) d x}{\left(3 x^{4}+9 x^{2}\right)^{5}}&= \frac{1}{6}\int \frac{\left(12 x^{3}+18 x\right) d x}{\left(3 x^{4}+9 x^{2}\right)^{5}}\\ &=\frac{1}{6} \left(12 x^{3}+18 x\right) (3 x^{4}+9 x^{2})^{-5}\\ &=-\frac{1}{24} (3 x^{4}+9 x^{2})^{-4}+c. \end{align*}
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