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: 26


$(3) \ln |\dfrac{\sqrt[3] {x}}{1+\sqrt[3] {x}}|+c$

Work Step by Step

Consider the integral $\int \dfrac{dx}{x(1+\sqrt[3] {x})}$ Let us take the help of the substitution method. $a=\sqrt[3] {x} \implies da=\dfrac{dx}{3x^{2/3}}; dx=3a^2 da$ The given integral can be re-written as: $\int \dfrac{dx}{x(1+\sqrt[3] {x})}=(3) \int \dfrac{da}{a(1+a)}$ This implies that $(3) \int \dfrac{da}{a(1+a)}=(3) \int(\dfrac{1}{a}- \dfrac{1}{(1+a)})da$ We use the formula: $\int x^n dx=\int\dfrac{x^{n+1}}{n+1} dx$ or, $(3)\ln | \dfrac{a}{1+a}|+C=(3) \ln |\dfrac{\sqrt[3] {x}}{1+\sqrt[3] {x}}|+c$
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