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: 43


$$\int \frac{x^{3}}{\sqrt[3]{x^{2}+1}} d x=\frac{3}{10}\left(x^{2}+1\right)^{5 / 3}-\frac{3}{4}\left(x^{2}+1\right)^{2 / 3}+C$$

Work Step by Step

Given$$\int \frac{x^{3}}{\sqrt[3]{x^{2}+1}} d x$$ \begin{array}{l}{\text { Let } u=\sqrt[3]{x^{2}+1}} \\ {\text { Then, } x=\sqrt{u^{3}-1}} \\ {d x=\frac{3 u^{2}}{2 \sqrt{u^{3}-1}}}\end{array} So, we have \begin{array}{l} {I=\int \frac{x^{3}}{\sqrt[3]{x^{2}+1}} d x}\\{=\int \frac{\left(u^{3}-1\right) \sqrt{u^{3}-1}}{u} \times \frac{3 u^{2}}{2 \sqrt{u^{3}-1}} d u} \\ {=\int \frac{\left(u^{3}-1\right)(3 u)}{2} d u} \\ {=\frac{1}{2} \int\left(3 u^{4}-3 u\right) d u} \\ {=\frac{1}{2}\left(\frac{3}{5} u^{5}-\frac{3}{2} u^{2}\right)+C} \\ {=\frac{3}{10} u^{5}-\frac{3}{4} u^{2}+C}\\{=\frac{3}{10}\left(x^{2}+1\right)^{5 / 3}-\frac{3}{4}\left(x^{2}+1\right)^{2 / 3}+C}\end{array}
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