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: 39

Answer

$$-\frac{1}{5} \ln |x-1|-\frac{1}{x-1}+\frac{1}{10} \ln \left|x^{2}+9\right|-\frac{4}{15} \tan ^{-1}\left(\frac{x}{3}\right)+C$$

Work Step by Step

Given $$\int \frac{10 d x}{(x-1)^{2}\left(x^{2}+9\right)} $$ Since \begin{aligned} \frac{10}{(x-1)^{2}\left(x^{2}+9\right)}&=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C x+D}{x^{2}+9}\\ &= \frac{A(x-1)\left(x^{2}+9\right)+B\left(x^{2}+9\right)+(C x+D)(x-1)^{2}}{(x-1)^{2}\left(x^{2}+9\right)}\\ 10&=A(x-1)\left(x^{2}+9\right)+B\left(x^{2}+9\right)+(C x+D)(x-1)^{2} \end{aligned} Then$$ \text{at } x =1 \ \ \ \ \to B=1 $$ By comparing the coefficients, we get \begin{aligned} A+C &=0 \ \ \ \ \text{ coefficient of } x^3\\ 1-A-2 C+D &=0\ \ \text{ coefficient of } x^2\\ 9 A+C-2 D &=0\ \ \text{ coefficient of } x \\ 9-9 A+D &=10\ \ \ \ \text{ coefficient of } x^0\\ \end{aligned} Then $$A=\frac{-1}{5},\ \ \ \ C=\frac{1}{5},\ \ \ \ D=\frac{-1}{5} $$ Hence \begin{aligned} \int \frac{10 d x}{(x-1)^{2}\left(x^{2}+9\right)} &=-\frac{1}{5} \int \frac{d x}{x-1}+\int \frac{d x}{(x-1)^{2}}+\frac{1}{5} \int \frac{x d x}{x^{2}+9}-\frac{4}{5} \int \frac{d x}{x^{2}+9} \\ &=-\frac{1}{5} \ln |x-1|-\frac{1}{x-1}+\frac{1}{10} \ln \left|x^{2}+9\right|-\frac{4}{15} \tan ^{-1}\left(\frac{x}{3}\right)+C \end{aligned}
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