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: 20

Answer

$$\frac{1}{3} \ln |x+1|-\frac{1}{3} \ln |x+4|+\frac{8}{(x+4)^{2}}+C $$

Work Step by Step

Given $$\int \frac{\left(x^{2}-8 x\right) d x}{(x+1)(x+4)^{3}}$$ Since \begin{align*} \frac{\left(x^{2}-8 x\right)}{(x+1)(x+4)^{3}}&=\frac{A}{(x+1)}+\frac{B}{(x+4)}+\frac{C}{(x+4)^{2}}+\frac{D}{(x+4)^{3}}\\ &=\frac{A(x+4)^{3}+B(x+1)(x+4)^{2}+C(x+1)(x+4)+D(x+1)}{(x+1)(x+4)^{3}}\\ x^{2}-8 x&=A(x+4)^{3}+B(x+1)(x+4)^{2}+C(x+1)(x+4)+D(x+1) \end{align*} Then \begin{align*} \text{at } x&= -1\ \ \ \ \to A=1/3 \\ \text{at } x&= -4\ \ \ \ \to D=-16\\ \text{at } x&= 0\ \ \ \ \to 4B+C=-4/3\\ \text{at } x&= 1\ \ \ \ \to 5B+C=-5/3\\ \end{align*} Hence $A=-1/3,\ \ C=0$ and \begin{aligned} \int \frac{\left(x^{2}-8 x\right) d x}{(x+1)(x+4)^{3}} &=\int \frac{\frac{1}{3} d x}{(x+1)}-\int \frac{\frac{1}{3} d x}{(x+4)}-\int \frac{16 d x}{(x+4)^{3}} \\ &=\frac{1}{3} \int \frac{d x}{(x+1)}-\frac{1}{3} \int \frac{d x}{(x+4)}-16 \int \frac{d x}{(x+4)^{3}} \\ &=\frac{1}{3} \ln |x+1|-\frac{1}{3} \ln |x+4|+\frac{8}{(x+4)^{2}}+C \end{aligned}
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