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: 15

Answer

$$2 \ln |x+3|-\ln |x+5|-\frac{2}{3} \ln |3 x-2|+C$$

Work Step by Step

Given $$\int \frac{\left(x^{2}+3 x-44\right) d x}{(x+3)(x+5)(3 x-2)}$$ Since \begin{align*} \frac{\left(x^{2}+3 x-44\right)}{(x+3)(x+5)(3 x-2)}&=\frac{A}{(x+3)}+\frac{B}{(x+5)}+\frac{C}{(3 x-2)}\\ &=\frac{A(x+5)(3 x-2)+B(x+3)(3 x-2)+C(x+3)(x+5)}{(x+3)(x+5)(3 x-2)}\\ x^{2}+3 x-44&=A(x+5)(3 x-2)+B(x+3)(3 x-2)+C(x+3)(x+5) \end{align*} Then \begin{align*} \text{at } x&=-3\ \ \ \ \ A=2 \\ \text{at } x&=-5\ \ \ \ \ B=-1\\ \text{at } x&=2/3\ \ \ \ \ B=-2 \end{align*} Hence \begin{aligned} \int \frac{\left(x^{2}+3 x-44\right) d x}{(x+3)(x+5)(3 x-2)} &=\int \frac{2 d x}{(x+3)}-\int \frac{d x}{(x+5)}-\int \frac{2 d x}{(3 x-2)} \\ &=2 \ln |x+3|-\ln |x+5|-\frac{2}{3} \ln |3 x-2|+C \end{aligned}
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