Calculus (3rd Edition)

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

Chapter 5 - The Integral - 5.7 Substitution Method - Exercises - Page 275: 34


$$\frac{-1}{6 \left(4 x^{3}+3 x^{2}\right) }+c$$

Work Step by Step

Given $$\int \frac{2 x^{2}+x}{\left(4 x^{3}+3 x^{2}\right)^{2}} d x$$ Let $$u=4 x^{3}+3 x^{2}\ \ \ \Rightarrow \ \ du=6(2x^2+x)dx $$ then \begin{align*} \int \frac{2 x^{2}+x}{\left(4 x^{3}+3 x^{2}\right)^{2}} d x&=\frac{1}{6}\int \frac{du}{u^{2}} \\ &=\frac{-1}{6u}+c\\ &=\frac{-1}{6 \left(4 x^{3}+3 x^{2}\right) }+c\\ \end{align*}
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