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: 17



Work Step by Step

Given $$ \int_{0}^{\infty} \frac{ d x}{(x+1)\sqrt{x}}$$ Let $$u=\sqrt{x} ,\ \ \to du=\frac{dx}{2\sqrt{x}} $$ Then \begin{align*} \int_{0}^{\infty} \frac{ d x}{(x+1)\sqrt{x}}&=\int_{0}^{\infty} \frac{2 d u}{u^{2}+1}\\ &=\int_{0}^{\infty} \frac{2 d u}{u^{2}+1}\\ &=\lim _{b \rightarrow \infty} \int_{0}^{b} \frac{2 \mathrm{du}}{u^{2}+1}\\ &=\lim _{b \rightarrow \infty}\left[2 \tan ^{-1} u\right]_{0}^{b}\\ &=\lim _{b \rightarrow \infty}\left(2 \tan ^{-1} b-2 \tan ^{-1} 0\right)\\ &=\pi \end{align*}
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