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: 24


$$-\frac{1}{9} \ln |x-4|+\frac{1}{3} \frac{(x-4)^{-1}}{(-1)}+\frac{1}{9} \ln |x-1|+C$$

Work Step by Step

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