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: 25


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

Work Step by Step

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