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: 14

Answer

$$\frac{1}{5} \ln |x-3|+\frac{1}{20} \ln |x+2|-\frac{1}{4} \ln |x-2|+C$$

Work Step by Step

Given $$\int \frac{d x}{(x-2)(x-3)(x+2)}$$ Since \begin{align*} \frac{1}{(x-2)(x-3)(x+2)}&=\frac{A}{(x-2)}+\frac{B}{(x-3)}+\frac{C}{(x+2)}\\ &=\frac{A(x-3)(x+2)+B(x-2)(x+2)+C(x-2)(x-3)}{(x-2)(x-3)(x+2)}\\ 1&=A(x-3)(x+2)+B(x-2)(x+2)+C(x-2)(x-3) \end{align*} Then \begin{align*} \text{at } x&=2\ \ \ \ \ A=\frac{-1}{4} \\ \text{at } x&=3\ \ \ \ \ B=\frac{1}{20}\\ \text{at } x&=-2\ \ \ \ \ B=\frac{1}{5} \end{align*} Hence \begin{align*} \int \frac{1}{(x-2)(x-3)(x+2)}dx&=\frac{-1}{4}\int\frac{1}{(x-2)}dx+\frac{1}{20}\int\frac{1}{(x-3)}dx+\frac{1}{5}\int\frac{1}{(x+2)}dx\\ &=\frac{1}{5} \ln |x-3|+\frac{1}{20} \ln |x+2|-\frac{1}{4} \ln |x-2|+C \end{align*}
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