University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Practice Exercises - Page 474: 21


$\dfrac{x^2}{2}+\dfrac{4}{3} \ln |x+2|+\dfrac{2}{3} \ln |x-1|+C$

Work Step by Step

Consider the integral $\int \dfrac{x^3+x^2}{x^2+x-2}dx$ The given integral can be re-written as: $\int \dfrac{x^3+x^2}{x^2+x-2}dx=\dfrac{x^3+x^2-2x+2x}{x^2+x-2}dx$ This implies that $\int \dfrac{x^3+x^2-2x}{x^2+x-2} dx+ \int \dfrac{2x}{x^2+x-2}dx=\int x dx+\int \dfrac{\frac{4}{3}}{x+2}+\dfrac{\frac{2}{3}}{x-1} dx$ We use the formula: $\int x^n dx=\int\dfrac{x^{n+1}}{n+1} dx$ $=\dfrac{x^2}{2}+\dfrac{4}{3} \ln |x+2|+\dfrac{2}{3} \ln |x-1|+C$
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