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: 17


$$3 \ln |x-1|-2 \ln |x+1|-\frac{5}{(x+1)}+C$$

Work Step by Step

Given $$\int \frac{\left(x^{2}+11 x\right) d x}{(x-1)(x+1)^{2}}$$ Since \begin{align*} \frac{\left(x^{2}+11 x\right) }{(x-1)(x+1)^{2}}&=\frac{A}{(x-1)}+\frac{B}{(x+1)}+\frac{C}{(x+1)^{2}}\\ &=\frac{A(x+1)^{2}+B \cdot(x-1)(x+1)+C(x-1)}{(x-1)(x+1)^{2}}\\ x^{2}+11 x&=A(x+1)^{2}+B \cdot(x-1)(x+1)+C(x-1) \end{align*} Then \begin{align*} \text{at } x&=1 \ \ \ \ \to A=3 \\ \text{at } x&= -1\ \ \ \ \to C=5\\ \text{at } x&= 0\ \ \ \ \to B=-2 \end{align*} Hence \begin{aligned} \int \frac{\left(x^{2}+11 x\right) d x}{(x-1)(x+1)^{2}} &=\int \frac{3 d x}{(x-1)}-\int \frac{2 d x}{(x+1)}+\int \frac{5 d x}{(x+1)^{2}} \\ &=3 \int \frac{d x}{(x-1)}-2 \int \frac{d x}{(x+1)}+5 \int(x+1)^{-2} d x\\ &=3 \ln |x-1|-2 \ln |x+1|-\frac{5}{(x+1)}+C \end{aligned}
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