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: 37



Work Step by Step

$$A=\int\frac{x}{\sqrt{1+x}}dx$$ We set $u=\sqrt{1+x}$, which means $x+1=u^2$ and $x=u^2-1$ Then $$du=\frac{(1+x)'}{2\sqrt{1+x}}dx=\frac{1}{2\sqrt{1+x}}dx$$ That means $$\frac{1}{\sqrt{1+x}}dx=2du$$ Therefore, $$A=2\int (u^2-1)du$$ $$A=2\Big(\frac{u^3}{3}-u\Big)+C$$ $$A=2\Big(\frac{(\sqrt{1+x})^3}{3}-\sqrt{1+x}\Big)+C$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.