Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.4 - Integration of Rational Functions by Partial Fractions. - 7.4 Exercises - Page 501: 45

Answer

$$\int \frac{1}{\sqrt{x}-\sqrt[3]{x}} d x =2 \sqrt{x}+3 \sqrt[3]{x}+6 \sqrt[6]{x}+6 \ln |\sqrt[6]{x}-1|+C $$

Work Step by Step

Given $$\int \frac{1}{\sqrt{x}-\sqrt[3]{x}} d x \ \ \ \ \ \ \ \ \ \ [\text { Hint: Substitute } u=\sqrt[6]{x} .]$$ \begin{array}{l}{\text { Let } u=\sqrt[6]{x} \text { . Then } x=u^{6}, \text { so } d x=6 u^{5} d u \\ \text { and } \sqrt{x}=u^{3}, \sqrt[3]{x}=u^{2} \text { . Thus, }} \\ {\qquad \begin{aligned} I=& \int \frac{d x}{\sqrt{x}-\sqrt[3]{x}} \\ &=\int \frac{6 u^{5} d u}{u^{3}-u^{2}}\\ &=6 \int \frac{u^{5}}{u^{2}(u-1)} d u\\ &=6 \int \frac{u^{3}}{u-1} d u \\ &=6 \int \frac{u^{3}-1+1}{u-1} d u \\ &=6 \int\left( \frac{u^{3}-1}{u-1}+\frac{ 1}{u-1}\right)d u \\ &=6 \int\left( \frac{(u -1)(u^2+u+1)}{u-1}+\frac{ 1}{u-1}\right)d u \\ &=6 \int\left(u^{2}+u+1+\frac{1}{u-1}\right) d u \\ &=6\left(\frac{1}{3} u^{3}+\frac{1}{2} u^{2}+u+\ln |u-1|\right)+C\\ &=2 \sqrt{x}+3 \sqrt[3]{x}+6 \sqrt[6]{x}+6 \ln |\sqrt[6]{x}-1|+C \end{aligned}}\end{array}
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