University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 51


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

Work Step by Step

$$A=\int x\tan^{-1}xdx$$ We set $u=\tan^{-1}x$ and $dv=xdx$ This means $du=\frac{1}{1+x^2}dx$ and $v=\frac{x^2}{2}$ Applying the formula $\int udv=uv-\int vdu$, we have $$A=\frac{x^2\tan^{-1}x}{2}-\frac{1}{2}\int\frac{x^2}{1+x^2}dx$$ $$A=\frac{x^2\tan^{-1}x}{2}-\frac{1}{2}\int\frac{x^2+1-1}{x^2+1}dx$$ $$A=\frac{x^2\tan^{-1}x}{2}-\frac{1}{2}\int\Big(1-\frac{1}{x^2+1}\Big)dx$$ $$A=\frac{x^2\tan^{-1}x}{2}-\frac{1}{2}\Big(\int1dx-\int\frac{1}{x^2+1}dx\Big)$$ $$A=\frac{x^2\tan^{-1}x}{2}-\frac{1}{2}\Big(x-\tan^{-1}x\Big)+C$$ $$A=\frac{1}{2}(x^2\tan^{-1}x-x+\tan^{-1}x)+C$$
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