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: 18

Answer

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

Work Step by Step

Given $$\int \frac{\left(4 x^{2}-21 x\right) d x}{(x-3)^{2}(2 x+3)}$$ Since \begin{align*} \frac{ 4 x^{2}-21 x }{(x-3)^{2}(2 x+3)}&=\frac{A}{(2 x+3)}+\frac{B}{(x-3)}+\frac{C}{(x-3)^{2}}\\ &= \frac{ A(x-3)^{2}+B \cdot(2 x+3)(x-3)+C(2 x+3)}{(x-3)^{2}(2 x+3)}\\ 4 x^{2}-21 x&= A(x-3)^{2}+B \cdot(2 x+3)(x-3)+C(2 x+3) \end{align*} Then \begin{align*} \text{at } x&=-3/2 \ \ \ \ \to A=2 \\ \text{at } x&= 3\ \ \ \ \to C=-3\\ \text{at } x&= 0\ \ \ \ \to B=1 \end{align*} Hence \begin{aligned} \int \frac{\left(4 x^{2}-21 x\right) d x}{(x-3)^{2}(2 x+3)} &=\int \frac{2 d x}{(2 x+3)}+\int \frac{d x}{(x-3)}-\int \frac{3 d x}{(x-3)^{2}}\\ &=\ln |2 x+3|+\ln |x-3|+\frac{3}{(x-3)}+C \end{aligned}
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