Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - 7.5 Strategy for Integration - 7.5 Exercises - Page 547: 21

Answer

$$x\tan^{-1}x-x+\tan^{-1}x+c$$

Work Step by Step

Given $$\int\tan^{-1}\sqrt{x}dx$$ Let $u^2=x \ \ \ \Rightarrow \ \ 2udu=dx$, then $$\int\tan^{-1}\sqrt{x}dx= 2\int u\tan^{-1}udu$$ Let \begin{align*} u&= \tan^{-1}u\ \ \ \ \ \ \ \ \ \ \ \ dv=2udu\\ du&=\frac{1}{1+u^2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v= u^2 \end{align*} Then \begin{align*} \int\tan^{-1}\sqrt{x}dx&=\int\tan^{-1}\sqrt{x}dx\\ &=u^2 \tan^{-1}u-\int \frac{u^2}{1+u^2}du\\ &=u^2 \tan^{-1}u-\int \frac{u^2+1-1}{1+u^2}du\\ &=u^2 \tan^{-1}u -\int 1-\frac{ 1}{1+u^2}du\\ &=u^2 \tan^{-1}u-u+\tan^{-1}u+c\\ &=x\tan^{-1}x-x+\tan^{-1}x+c \end{align*}
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