Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 39

Answer

The answer is $\int \frac{x}{\sqrt{x+1}} dx = 2x (x+1)^{\frac{1}{2}} - \frac{4}{3} (x+1)^{\frac{3}{2}} + c$.

Work Step by Step

$\int \frac{x}{\sqrt{x+1}} dx = \int x (x+1)^{\frac{-1}{2}} dx$, and for $u=x$, then $du = dx$ and for $dv= (x+1)^{\frac{-1}{2}} dx$, then $v = 2 (x+1)^{\frac{1}{2}}$. Hence, integration by parts $\int u dv = u v - \int v du$ implies that $\int \frac{x}{\sqrt{x+1}} dx = x \times 2 (x+1)^{\frac{1}{2}} - \int 2 (x+1)^{\frac{1}{2}} dx = 2x (x+1)^{\frac{1}{2}} - \frac{4}{3} (x+1)^{\frac{3}{2}} + c$.
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