Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.8 - Improper Integrals - Exercises 8.8 - Page 501: 20


$$2 \pi^{2} $$

Work Step by Step

\begin{align*} \int_{0}^{\infty} \frac{16 \tan ^{-1} x}{1+x^{2}} d x&= 16\int_{0}^{\infty} \frac{1}{1+x^{2}} \tan ^{-1} x dx\\ &=\lim _{b \rightarrow \infty}\left[8\left(\tan ^{-1} x\right)^{2}\right]_{0}^{b}\\ &=\lim _{b \rightarrow \infty}\left[8\left(\tan ^{-1} b\right)^{2}-8\left(\tan ^{-1} 0\right)^{2}\right]\\&=8\left(\frac{\pi}{2}\right)^{2}-8(0)\\ &=2 \pi^{2} \end{align*}
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