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: 21

Answer

$$\ln |x|-\ln |x+2|+ \frac{2}{(x+2)}+ \frac{2}{(x+2)^{2}}+C$$

Work Step by Step

Given $$\int \frac{8 d x}{x(x+2)^{3}}$$ Since \begin{align*} \frac{8}{x(x+2)^{3}}&=\frac{A}{x}+\frac{B}{(x+2)}+\frac{C}{(x+2)^{2}}+\frac{D}{(x+2)^{3}}\\ &= \frac{A(x+2)^{3}+B x(x+2)^{2}+C x(x+2)+D x}{x(x+2)^{3}}\\ 8&= A(x+2)^{3}+B x(x+2)^{2}+C x(x+2)+D x \end{align*} Then \begin{align*} \text{at } x&= 0\ \ \ \ \to A=1 \\ \text{at } x&= -2\ \ \ \ \to D=-4\\ \text{at } x&= 1\ \ \ \ \to 3B+C=-5\\ \text{at } x&= - 1\ \ \ \ \to -B-C=3\\ \end{align*} Hence $B=-1,\ \ \ C= -2$ and \begin{aligned} \int \frac{8 d x}{x(x+2)^{3}} &=\int \frac{d x}{x}-\int \frac{d x}{(x+2)}-\int \frac{2 d x}{(x+2)^{2}}-\int \frac{4 d x}{(x+2)^{3}} \\ &=\int \frac{d x}{x}-\int \frac{d x}{(x+2)}-2 \int(x+2)^{-2} d x-4 \int(x+2)^{-3} d x\\ &=\ln |x|-\ln |x+2|+\frac{2}{(x+2)}+ \frac{2}{(x+2)^{2}}+C \end{aligned}
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