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: 28

Answer

$$\frac{11}{3} \ln |x|-\frac{2}{x}-\frac{9}{2} \ln |x-1|+\frac{5}{6} \ln |x-3|+C$$

Work Step by Step

Given $$\int \frac{3 x+6}{x^{2}(x-1)(x-3)} d x$$ Since \begin{aligned} \frac{3 x+6}{x^{2}(x-1)(x-3)} &=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{(x-1)}+\frac{D}{(x-3)}\\ &=\frac{ A x(x-1)(x-3)+B(x-1)(x-3) +C x^{2}(x-3)+D x^{2}(x-1)}{x^{2}(x-1)(x-3)}\\ 3 x+6&= A x(x-1)(x-3)+B(x-1)(x-3) +C x^{2}(x-3)+D x^{2}(x-1) \ \end{aligned} Then \begin{align*} \text{at } x&=1 \ \ \ \ \to C=\frac{-9}{2}\\ \text{at } x&=0\ \ \ \ \to B=2\\ \text{at } x&= 3\ \ \ \ \to D= \frac{5}{6}\\ \text{at } x&=-1\ \ \ \ \to A= \frac{11}{3} \end{align*} Hence \begin{aligned} \int \frac{(3 x+6) d x}{x^{2}(x-1)(x-3)}&=\int \frac{\frac{11}{3} d x}{x}+\int \frac{2 d x}{x^{2}}-\int \frac{\frac{9}{2} d x}{(x-1)}+\int \frac{\frac{5}{6} d x}{(x-3)}\\ &=\frac{11}{3} \ln |x|-\frac{2}{x}-\frac{9}{2} \ln |x-1|+\frac{5}{6} \ln |x-3|+C \end{aligned}
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