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: 26


$$\frac{1}{9} \ln |x+1|-\frac{1}{9} \ln |x-2|-\frac{1}{3} \frac{1}{x-2}+C$$

Work Step by Step

Given $$\int \frac{d x}{x^{3}-3 x^{2}+4}$$ Since \begin{aligned} \frac{1}{x^{3}-3 x+4} &=\frac{1}{(x+1)(x-2)^{2}} \\ &=\frac{\mathrm{A}}{x+1}+\frac{\mathrm{B}}{x-2}+\frac{\mathrm{C}}{(x-2)^{2}} \\ &=\frac{(x-2)^{2} \mathrm{A}+\mathrm{B}(x+1)(x-2)+\mathrm{C}(x+1)}{x^{3}-3 x+4} \\ 1 &=(x-2)^{2} \mathrm{A}+\mathrm{B}(x+1)(x-2)+\mathrm{C}(x+1) \end{aligned} Then \begin{align*} \text{at } x&=-1\ \ \ \ \to A= \frac{1}{9} \\ \text{at } x&=2\ \ \ \ \to B=\frac{-1}{9}\\ \text{at } x&=0\ \ \ \ \to C=\frac{1}{3} \end{align*} Hence \begin{aligned} \int \frac{d x}{x^{3}-3 x+4} &=\frac{1}{9} \int \frac{1}{x+1} d x-\frac{1}{9} \int \frac{1}{x-2} d x+\frac{1}{3} \int \frac{1}{(x-2)^{2}} d x\\ &=\frac{1}{9} \ln |x+1|-\frac{1}{9} \ln |x-2|-\frac{1}{3} \frac{1}{x-2}+C \end{aligned}
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