Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.8 - Improper Integrals - 7.8 Exercises - Page 534: 17

Answer

Converges to $\ln 2$

Work Step by Step

Type I. $I=\displaystyle \int_{1}^{\infty}\frac{1}{x^{2}+x}dx=\lim_{t\rightarrow\infty}\int_{1}^{t}\frac{1}{x(x+1)}dx$ $\displaystyle \frac{1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}$ $\displaystyle \frac{1}{x(x+1)}=\frac{A(x+1)+Bx}{x}$ $\displaystyle \frac{1}{x(x+1)}=\frac{(A+B)x+A}{x}\Rightarrow\left\{\begin{array}{ll} A=1 & \\ A+B=0 & \Rightarrow B=-1 \end{array}\right\}$ $I=\displaystyle \lim_{t\rightarrow\infty}\int_{1}^{t}(\frac{1}{x}-\frac{1}{x+1})dx\quad $ $=\displaystyle \lim_{t\rightarrow\infty}[\ln|x|-\ln|x+1|]_{1}^{t}$ $=\displaystyle \lim_{t\rightarrow\infty}\left[\ln\left|\frac{x}{x+1}\right|\right]_{1}^{t}$ $=\displaystyle \lim_{t\rightarrow\infty}(\ln\frac{t}{t+1}-\ln\frac{1}{2})$ $=0-\displaystyle \ln\frac{1}{2}$ $=\ln 2$ (converges)
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