University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 46


$$\int (x+5)(x-5)^{1/3}dx=\frac{3}{7}(x-5)^{7/3}+\frac{15}{2}(x-5)^{4/3}+C$$

Work Step by Step

$$A=\int (x+5)(x-5)^{1/3}dx$$ We set $u=x-5$, which means $x+5=u+10$ Then $$du=dx$$ Therefore, $$A=\int(u+10)u^{1/3}du=\int\Big(u^{4/3}+10u^{1/3}\Big)du$$ $$A=\frac{u^{7/3}}{\frac{7}{3}}+\frac{10u^{4/3}}{\frac{4}{3}}+C$$ $$A=\frac{3u^{7/3}}{7}+\frac{15u^{4/3}}{2}+C$$ $$A=\frac{3}{7}(x-5)^{7/3}+\frac{15}{2}(x-5)^{4/3}+C$$
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