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: 33


$$\ln 2$$

Work Step by Step

\begin{align*} \int_{-1}^{\infty} \frac{d \theta}{\theta^{2}+5 \theta+6}&=\int_{-1}^{\infty} \frac{d \theta}{ \theta+2}-\int_{-1}^{\infty} \frac{d \theta}{\theta+3}\\ &=\lim _{b \rightarrow \infty}\int_{-1}^{b} \frac{d \theta}{ \theta+2}-\lim _{b \rightarrow \infty}\int_{-1}^{b} \frac{d \theta}{\theta+3}\\ &=\lim _{b \rightarrow \infty}\left[\ln {|\theta+2|}-\ln {|\theta+3|} \right]\bigg|_{-1}^{b}\\ &=\lim _{b \rightarrow \infty}\left[\ln \left|\frac{\theta+2}{\theta+3}\right|\right]\bigg|_{-1}^{b}\\ &=\lim _{b \rightarrow \infty}\left[\ln \left|\frac{b+2}{b+3}\right|-\ln \left|\frac{-1+2}{-1+3}\right|\right]\\ &=0-\ln \left(\frac{1}{2}\right)=\ln 2 \end{align*}
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