Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 461: 82


$$\ln \left|\frac{6}{5}\right|$$

Work Step by Step

Given $$\int_{1}^{\infty} \frac{d x}{(x+2)(2 x+3)}$$ Since \begin{aligned} \frac{1}{(x+2)(2 x+3)} &=\frac{A}{(x+2)}+\frac{B}{(2 x+3)} \\ 1 &=A(2 x+3)+B(x+2) \end{aligned} At $x=-2$, $B= 2$, and at $x= -3/2$ $A= -1$. Thus: \begin{aligned} \int_{1}^{\infty} \frac{d x}{(x+2)(2 x+3)}&=\lim_{R\to\infty}\int_{1}^{R} \frac{d x}{(x+2)(2 x+3)} \\ &=\lim_{R\to\infty}\int_{1}^{R}\left[\frac{2}{(2 x+3)}-\frac{1}{(x+2)}\right] d x \\ &=\lim_{R\to\infty}[\ln |2 x+3|-\ln |x+2|]_{1}^{R}\\ &=\lim_{R\to\infty}\ln \left|\frac{2 R+3}{R+2}\right|-\ln \left|\frac{5}{3}\right|\\ &=\ln |2|-\ln \left|\frac{5}{3}\right|\\ &=\ln \left|\frac{6}{5}\right| \end{aligned}
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