Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - Review - Exercises - Page 578: 14


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

Work Step by Step

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