Calculus 8th Edition

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

Chapter 7 - Techniques of Integration - 7.4 Integration of Rational Functions by Partial Fractions - 7.4 Exercises - Page 541: 54


$$\int x \tan ^{-1} x=\frac{\left(x^{2}+1\right) \tan ^{-1} x}{2}-\frac{x}{2}+C $$

Work Step by Step

Given $$\int x \tan ^{-1} x$$ So by partition technique Let $$u=\tan ^{-1} x \Rightarrow du= \frac{1}{1+x^2}dx$$ $$dv=x dx \Rightarrow v= \frac{x^2}{2} $$ So, we get \begin{aligned} I&=uv- \int vdu\\ &=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int \frac{x^{2}}{1+x^{2}} d x\\ &=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int \frac{x^{2}+1}{1+x^{2}}-\frac{1}{1+x^{2}} d x\\ & =\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int 1-\frac{1}{1+x^{2}} d x\\ &=\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+C\\ &=\frac{\left(x^{2}+1\right) \tan ^{-1} x}{2}-\frac{x}{2}+C \end{aligned}
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