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: 38


$$2 \ln |x-2|+4 \ln |x+2|+\frac{1}{x+2}+C$$

Work Step by Step

Given $$\int \frac{6 x^{2}+7 x-6}{\left(x^{2}-4\right)(x+2)} d x$$ Since \begin{aligned} \frac{6 x^{2}+7 x-6}{\left(x^{2}-4\right)(x+2)} &=\frac{6 x^{2}+7 x-6}{(x-2)(x+2)(x+2)} \\ &=\frac{6 x^{2}+7 x-6}{(x-2)(x+2)^{2}} \\ &=\frac{A}{(x-2)}+\frac{B}{(x+2)}+\frac{C}{(x+2)^{2}} \\ 6 x^{2}+7 x-6 &=A(x+2)^{2}+B(x-2)(x+2)+C(x-2) \end{aligned} Then\begin{align*} \text{at } x&=2 \ \ \ \ \to A=2 \\ \text{at } x&= -2\ \ \ \ \to C=-1\\ \text{at } x&= 0\ \ \ \ \to B=4 \end{align*} Hence \begin{aligned} \int \frac{6 x^{2}+7 x-6}{\left(x^{2}-4\right)(x+2)} d x &=\int \frac{2}{x-2} d x+\int \frac{4}{x+2} d x-\int \frac{1}{(x+2)^{2}} d x \\ &=\int \frac{2}{x-2} d x+\int \frac{4}{x+2} d x-\int(x+2)^{-2} d x\\ &=2 \ln |x-2|+4 \ln |x+2|+\frac{1}{x+2}+C \end{aligned}
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