Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.5 The Method of Partial Fractions - Exercises - Page 423: 11


$$\ln x-\ln |2x+1|+C$$

Work Step by Step

Given $$\int \frac{d x}{x(2 x+1)}$$ Since \begin{align*} \frac{1}{x(2 x+1)}&=\frac{A}{x}+\frac{B}{2x+1}\\ &=\frac{A(2x+1)+Bx}{x(2x+1)}\\ 1&=A(2x+1)+Bx \end{align*} Then \begin{align*} \text{at } x&=0\ \ \ \ \ A=1\\ \text{at } x&=\frac{-1}{2}\ \ \ \ \ B=-2 \end{align*} Hence \begin{align*} \int \frac{1}{x(2 x+1)}dx&=\int \frac{1}{x}dx+\int \frac{-2}{2x+1}dx\\ &=\ln x-\ln |2x+1|+C \end{align*}
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