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: 37


$$6 x-14 \ln |x+3|+2 \ln |x-1|+C$$

Work Step by Step

Given $$\int \frac{\left(6 x^{2}+2\right) d x}{x^{2}+2 x-3}$$ By using long division, we get: \begin{align*} \frac{6 x^{2}+2}{x^{2}+2 x-3}&=6-\frac{12 x-20}{x^{2}+2 x-3}\\ &=6-\frac{12 x-20}{(x+3)(x-1)} \end{align*} and \begin{align*} \frac{12 x-20}{(x+3)(x-1)}&=\frac{A}{x+3}+\frac{B}{x-1}\\ &=\frac{A(x-1)+B(x+3)}{(x+3)(x-1)}\\ 12 x-20&=A(x-1)+B(x+3) \end{align*} Then\begin{align*} \text{at } x&=-3 \ \ \ \ \to A=14 \\ \text{at } x&= 1\ \ \ \ \to B=-2 \end{align*} Hence \begin{aligned} \int \frac{6 x^{2}+2}{x^{2}+2 x-3} &=\int 6-\frac{14}{x+3}+\frac{2}{x-1} d x\\ &=\int 6 \, d x-14 \int \frac{1}{x+3} d x+2 \int \frac{1}{x-1} d x \\ &=6 x-14 \ln |x+3|+2 \ln |x-1|+C \end{aligned}
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