Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.5 - Strategy for Integration - 7.5 Exercises - Page 507: 21

Answer

$(x+1) \arctan\sqrt{x}-\sqrt{x}+C$

Work Step by Step

Strategy: see if we can lose the square root...$\left[\begin{array}{ll} t=\sqrt{x} & t^{2}=x\\ & 2tdt=dx \end{array}\right]$ $I= \displaystyle \int \arctan \displaystyle \sqrt{x}dx=\int \arctan t(2tdt)= \displaystyle \int u(dv),\qquad $...by parts $\left[\begin{array}{ll} u= \arctan t & dv=2tdt\\ du=\dfrac{1}{1+t^{2}}dt & v=t^{2} \end{array}\right]$ $=uv-\displaystyle \int vdu$ $=t^{2} \displaystyle \arctan t-\int\frac{t^{2}}{1+t^{2}}dt$ $=t^{2} \displaystyle \arctan t-\int\frac{1+t^{2}-1}{1+t^{2}}dt$ $=t^{2} \displaystyle \arctan t-\int(1-\frac{1}{1+t^{2}})dt$ $=t^{2} \displaystyle \arctan t-\int dt+\int\frac{1}{1+t^{2}}dt$ both integrals are in the table: 1. and 17. $=t^{2}\arctan t-t+ \arctan t+C$ $=(t^{2}+1) \arctan t-t+C$ ... bring x back $=(x+1) \arctan\sqrt{x}-\sqrt{x}+C$
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