Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 42


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

Work Step by Step

Given $$\int \frac{\left(7 x^{2}+x\right) d x}{(x-2)(2 x+1)(x+1)}$$ Since \begin{aligned} \frac{\left(7 x^{2}+x\right)}{(x-2)(2 x+1)(x+1)} &=\frac{A}{(x-2)}+\frac{B}{(2 x+1)}+\frac{C}{(x+1)} \\ &=\frac{A(2 x+1)(x+1)+B(x-2)(x+1)+C(x-2)(2 x+1)}{(x-2)(2 x+1)(x+1)}\\ 7 x^{2}+x &=A(2 x+1)(x+1)+B(x-2)(x+1)+C(x-2)(2 x+1)\\ \end{aligned} Then \begin{align*} \text{at } x&=2 \ \ \ \ \to A=2 \\ \text{at } x&=-1/2\ \ \ \ \to B=-1 \\ \text{at } x&=-1\ \ \ \ \to C=2 \end{align*} Hence \begin{aligned} \int \frac{\left(7 x^{2}+x\right)}{(x-2)(2 x+1)(x+1)} d x &=\int \frac{2}{(x-2)} d x-\int \frac{1}{(2 x+1)} d x+\int \frac{2}{(x+1)} d x \\ &=2 \ln |x-2|-\frac{1}{2} \ln |2 x+1|+2 \ln |x+1|+C \end{aligned}
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